回答編集履歴
7
バグ修正
test
CHANGED
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@@ -218,9 +218,13 @@
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218
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}
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+
for (; i < N; i++)
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-
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if (a1[i] != t) b[k++] = t = a1[i];
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-
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for (; j < N; j++)
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-
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227
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if (a2[j] != t) b[k++] = t = a2[j];
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228
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return k;
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6
コードの改善
test
CHANGED
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@@ -212,16 +212,12 @@
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212
212
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213
213
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else {
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214
214
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215
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-
if (a1[i] == a2[j]) i++;
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216
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-
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217
215
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if (a2[j] != t) b[k++] = t = a2[j];
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218
216
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217
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j++;
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218
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}
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220
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223
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-
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224
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-
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225
221
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while (i < N) b[k++] = a1[i++];
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226
222
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227
223
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while (j < N) b[k++] = a2[j++];
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5
コードの改善
test
CHANGED
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@@ -210,7 +210,9 @@
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210
210
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211
211
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}
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212
212
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213
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else {
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214
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213
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-
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215
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if (a1[i] == a2[j]) i++;
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214
216
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215
217
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if (a2[j] != t) b[k++] = t = a2[j];
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218
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@@ -218,13 +220,7 @@
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}
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222
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-
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223
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+
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222
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-
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223
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-
if (a1[i] != t) b[k++] = t = a1[i];
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224
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-
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225
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-
i++, j++;
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226
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-
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227
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-
}
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228
224
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229
225
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while (i < N) b[k++] = a1[i++];
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230
226
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4
ソートを使うコードを追加
test
CHANGED
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@@ -111,3 +111,179 @@
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111
111
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ビット演算を使わなければ理解できるのでしょうか？
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112
112
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113
113
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コメントをお願いします。
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114
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+
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115
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116
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117
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+
**追記2**
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118
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119
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+
ソートした 2つの配列を先頭から見ていって、積集合と和集合を求めるやり方。
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120
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+
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121
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+
```C
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122
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+
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123
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+
#include <stdio.h>
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124
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125
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126
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+
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127
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+
#define N 10 // 配列の要素数
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128
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+
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129
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#define A 1 // 要素の値の範囲 A以上 B以下
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130
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131
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#define B 10
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132
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133
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134
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135
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+
void print(int a[], int n)
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136
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137
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{
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138
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139
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+
for (int i = 0; i < n; i++)
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140
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+
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141
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+
printf(" %d", a[i]);
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142
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+
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143
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putchar('\n');
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144
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+
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145
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}
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146
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+
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147
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148
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149
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void sort(int a[])
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150
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151
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{
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152
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153
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+
for (int i = 0; i < N-1; i++)
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154
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+
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155
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+
for (int j = i+1; j < N; j++)
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156
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+
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157
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+
if (a[i] > a[j]) {
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158
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+
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159
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+
int t = a[i]; a[i] = a[j]; a[j] = t;
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160
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+
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161
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}
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162
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163
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}
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164
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+
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165
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166
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+
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167
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+
int intersection(int a1[], int a2[], int b[])
|
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168
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+
|
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169
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+
{
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170
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+
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171
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+
int i = 0, j = 0, k = 0, t = A-1;
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172
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173
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+
while (i < N && j < N)
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174
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+
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175
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+
if (a1[i] < a2[j])
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176
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+
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177
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+
i++;
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178
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+
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179
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+
else if (a1[i] > a2[j])
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180
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+
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181
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+
j++;
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182
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+
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183
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+
else {
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184
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+
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185
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+
if (a1[i] != t) b[k++] = t = a1[i];
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186
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+
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187
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+
i++, j++;
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188
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+
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189
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+
}
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190
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+
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191
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+
return k;
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192
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+
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193
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+
}
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194
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+
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195
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+
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196
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+
|
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197
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+
int union_set(int a1[], int a2[], int b[])
|
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198
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+
|
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199
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+
{
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200
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+
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201
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+
int i = 0, j = 0, k = 0, t = A-1;
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202
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+
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203
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+
while (i < N && j < N)
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204
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+
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205
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+
if (a1[i] < a2[j]) {
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206
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+
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207
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+
if (a1[i] != t) b[k++] = t = a1[i];
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208
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+
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209
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+
i++;
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210
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+
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211
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+
}
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212
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+
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213
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+
else if (a1[i] > a2[j]) {
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214
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+
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215
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+
if (a2[j] != t) b[k++] = t = a2[j];
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216
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+
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217
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j++;
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218
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+
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219
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+
}
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220
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+
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221
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+
else {
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222
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+
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223
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+
if (a1[i] != t) b[k++] = t = a1[i];
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224
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+
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225
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+
i++, j++;
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226
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+
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227
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}
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228
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+
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229
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+
while (i < N) b[k++] = a1[i++];
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230
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+
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231
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+
while (j < N) b[k++] = a2[j++];
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232
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+
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233
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+
return k;
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234
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+
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235
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+
}
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236
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+
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237
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+
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238
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+
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239
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+
int main(void)
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240
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+
|
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241
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+
{
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242
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+
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243
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+
int a1[N] = { 5, 8, 4, 3, 7, 9, 7, 9, 5, 1 };
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244
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+
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245
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+
int a2[N] = { 7, 6, 1, 5, 9, 9, 10, 8, 4, 3 };
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|
246
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+
|
|
247
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+
printf("配列1:"), print(a1, N);
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248
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+
|
|
249
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+
printf("配列2:"), print(a2, N);
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|
250
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+
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251
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+
sort(a1);
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252
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+
|
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253
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+
printf("配列1(昇順):"), print(a1, N);
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254
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+
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255
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+
sort(a2);
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256
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+
|
|
257
|
+
printf("配列2(昇順):"), print(a2, N);
|
|
258
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+
|
|
259
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+
int b[B-A+1], n;
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260
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+
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261
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+
n = intersection(a1, a2, b);
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|
262
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+
|
|
263
|
+
printf("共通する数:"), print(b, n);
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|
264
|
+
|
|
265
|
+
n = union_set(a1, a2, b);
|
|
266
|
+
|
|
267
|
+
printf("どちらかにある数:"), print(b, n);
|
|
268
|
+
|
|
269
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+
}
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|
270
|
+
|
|
271
|
+
```
|
|
272
|
+
|
|
273
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+
実行結果
|
|
274
|
+
|
|
275
|
+
```text
|
|
276
|
+
|
|
277
|
+
配列1: 5 8 4 3 7 9 7 9 5 1
|
|
278
|
+
|
|
279
|
+
配列2: 7 6 1 5 9 9 10 8 4 3
|
|
280
|
+
|
|
281
|
+
配列1(昇順): 1 3 4 5 5 7 7 8 9 9
|
|
282
|
+
|
|
283
|
+
配列2(昇順): 1 3 4 5 6 7 8 9 9 10
|
|
284
|
+
|
|
285
|
+
共通する数: 1 3 4 5 7 8 9
|
|
286
|
+
|
|
287
|
+
どちらかにある数: 1 3 4 5 6 7 8 9 10
|
|
288
|
+
|
|
289
|
+
```
|
3
コードの修正
test
CHANGED
|
@@ -88,7 +88,7 @@
|
|
|
88
88
|
|
|
89
89
|
int a2[N] = { 7, 6, 1, 5, 9, 9, 10, 8, 4, 3 };
|
|
90
90
|
|
|
91
|
-
int b1[B] = { 0 }, b2[B] = { 0 }, c[B+1] = { 0 }, d[B+1] = { 0 };
|
|
91
|
+
int b1[B+1] = { 0 }, b2[B+1] = { 0 }, c[B+1] = { 0 }, d[B+1] = { 0 };
|
|
92
92
|
|
|
93
93
|
for (int i = 0; i < N; i++) b1[a1[i]] = b2[a2[i]] = 1;
|
|
94
94
|
|
2
ビット演算を使わないコードを追加
test
CHANGED
|
@@ -47,3 +47,67 @@
|
|
|
47
47
|
}
|
|
48
48
|
|
|
49
49
|
```
|
|
50
|
+
|
|
51
|
+
**追記**
|
|
52
|
+
|
|
53
|
+
ビット演算を使わないやり方です。
|
|
54
|
+
|
|
55
|
+
```C
|
|
56
|
+
|
|
57
|
+
#include <stdio.h>
|
|
58
|
+
|
|
59
|
+
|
|
60
|
+
|
|
61
|
+
#define N 10 // 配列の要素数
|
|
62
|
+
|
|
63
|
+
#define A 1 // 要素の値の範囲 A以上 B以下
|
|
64
|
+
|
|
65
|
+
#define B 10
|
|
66
|
+
|
|
67
|
+
|
|
68
|
+
|
|
69
|
+
void print(int b[B+1])
|
|
70
|
+
|
|
71
|
+
{
|
|
72
|
+
|
|
73
|
+
for (int i = A; i <= B; i++)
|
|
74
|
+
|
|
75
|
+
if (b[i]) printf(" %d", i);
|
|
76
|
+
|
|
77
|
+
putchar('\n');
|
|
78
|
+
|
|
79
|
+
}
|
|
80
|
+
|
|
81
|
+
|
|
82
|
+
|
|
83
|
+
int main(void)
|
|
84
|
+
|
|
85
|
+
{
|
|
86
|
+
|
|
87
|
+
int a1[N] = { 5, 8, 4, 3, 7, 9, 7, 9, 5, 1 };
|
|
88
|
+
|
|
89
|
+
int a2[N] = { 7, 6, 1, 5, 9, 9, 10, 8, 4, 3 };
|
|
90
|
+
|
|
91
|
+
int b1[B] = { 0 }, b2[B] = { 0 }, c[B+1] = { 0 }, d[B+1] = { 0 };
|
|
92
|
+
|
|
93
|
+
for (int i = 0; i < N; i++) b1[a1[i]] = b2[a2[i]] = 1;
|
|
94
|
+
|
|
95
|
+
for (int i = A; i <= B; i++) {
|
|
96
|
+
|
|
97
|
+
c[i] = b1[i] * b2[i]; // 積集合
|
|
98
|
+
|
|
99
|
+
d[i] = b1[i] + b2[i]; // 和集合
|
|
100
|
+
|
|
101
|
+
}
|
|
102
|
+
|
|
103
|
+
printf("共通する数:"), print(c);
|
|
104
|
+
|
|
105
|
+
printf("どちらかにある数:"), print(d);
|
|
106
|
+
|
|
107
|
+
}
|
|
108
|
+
|
|
109
|
+
```
|
|
110
|
+
|
|
111
|
+
ビット演算を使わなければ理解できるのでしょうか？
|
|
112
|
+
|
|
113
|
+
コメントをお願いします。
|
1
10の意味の違いが分かるようにコードを修正
test
CHANGED
|
@@ -6,11 +6,19 @@
|
|
|
6
6
|
|
|
7
7
|
|
|
8
8
|
|
|
9
|
+
#define N 10 // 配列の要素数
|
|
10
|
+
|
|
11
|
+
#define A 1 // 要素の値の範囲 A以上 B以下
|
|
12
|
+
|
|
13
|
+
#define B 10
|
|
14
|
+
|
|
15
|
+
|
|
16
|
+
|
|
9
17
|
void print(int b)
|
|
10
18
|
|
|
11
19
|
{
|
|
12
20
|
|
|
13
|
-
for (int i =
|
|
21
|
+
for (int i = A; i <= B; i++)
|
|
14
22
|
|
|
15
23
|
if (b >> i & 1) printf(" %d", i);
|
|
16
24
|
|
|
@@ -24,13 +32,13 @@
|
|
|
24
32
|
|
|
25
33
|
{
|
|
26
34
|
|
|
27
|
-
int a1[] = { 5, 8, 4, 3, 7, 9, 7, 9, 5, 1 };
|
|
35
|
+
int a1[N] = { 5, 8, 4, 3, 7, 9, 7, 9, 5, 1 };
|
|
28
36
|
|
|
29
|
-
int a2[] = { 7, 6, 1, 5, 9, 9, 10, 8, 4, 3 };
|
|
37
|
+
int a2[N] = { 7, 6, 1, 5, 9, 9, 10, 8, 4, 3 };
|
|
30
38
|
|
|
31
39
|
int b1 = 0, b2 = 0;
|
|
32
40
|
|
|
33
|
-
for (int i = 0; i <
|
|
41
|
+
for (int i = 0; i < N; i++) b1 |= 1 << a1[i], b2 |= 1 << a2[i];
|
|
34
42
|
|
|
35
43
|
printf("共通する数:"), print(b1 & b2);
|
|
36
44
|
|