回答編集履歴
2
変数名の修正
test
CHANGED
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@@ -146,15 +146,15 @@
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146
146
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147
147
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for(int j = 0; j < MAX_NUMBER; j++) {
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148
148
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149
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-
//n
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149
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+
//n行ごとの塊が幾つ目か
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150
150
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151
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-
int
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151
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+
int lines = j / n;
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152
152
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153
153
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154
154
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155
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-
//行が偶数番目か奇数番目かを判定
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155
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+
//行の塊が偶数番目か奇数番目かを判定
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156
156
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157
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-
if(
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157
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+
if(lines % 2 == 0 ){
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158
158
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159
159
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System.out.println(patternA);
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160
160
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1
別パターン
test
CHANGED
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@@ -73,3 +73,101 @@
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73
73
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```
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74
74
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75
75
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ifで使っている`A^B`は排他的論理和といい，AとBの一方がtrue,他方がfalseの時にtrueを返します．
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76
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+
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77
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+
---
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78
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+
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79
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+
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80
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+
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81
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+
あるいは，列ごとに表示する文字列のパターンは2通りしかないので，それを先に作ってしまうのも手です．
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82
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+
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83
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+
```lang-java
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84
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+
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85
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+
import java.util.Scanner;
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86
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+
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87
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+
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88
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+
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89
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public class Amimemoyou {
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90
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91
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public static void main (String[] args) {
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92
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93
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final int MAX_NUMBER = 50;
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94
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95
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Scanner stdin = new Scanner (System.in);
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96
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+
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97
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+
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98
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+
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99
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System.out.print("Size> ");
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100
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101
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int n = stdin.nextInt();
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102
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+
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103
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System.out.print("\n");
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104
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+
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105
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if(n >= MAX_NUMBER) return;
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106
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+
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107
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+
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108
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+
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109
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StringBuilder lineA = new StringBuilder(),
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110
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111
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lineB = new StringBuilder();
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112
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113
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//1行(50文字)分の文字列を作る
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114
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+
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115
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for(int i = 0; i < MAX_NUMBER; i++) {
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116
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+
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117
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//n文字ごとの塊で幾つ目か
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118
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+
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119
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int columns = i / n;
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120
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+
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121
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if(columns % 2 == 0){
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122
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+
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123
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lineA.append(" ");
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124
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+
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125
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lineB.append("*");
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126
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+
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127
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} else {
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128
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+
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129
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lineA.append("*");
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130
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+
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131
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lineB.append(" ");
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132
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+
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133
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}
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134
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+
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135
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}
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136
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+
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137
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+
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138
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+
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139
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String patternA = lineA.toString(),
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140
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+
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141
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patternB = lineB.toString();
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142
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+
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143
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+
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144
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+
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145
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//最大で50行
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146
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+
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147
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for(int j = 0; j < MAX_NUMBER; j++) {
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148
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+
|
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149
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+
//n文字ごとの塊が幾つ目か
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150
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+
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151
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int columns = j / n;
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152
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+
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153
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+
|
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154
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+
|
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155
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+
//行が偶数番目か奇数番目かを判定
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156
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+
|
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157
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if(columns % 2 == 0 ){
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158
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+
|
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159
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System.out.println(patternA);
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160
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+
|
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161
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} else {
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162
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+
|
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163
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+
System.out.println(patternB);
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164
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+
|
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165
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}
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166
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+
|
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167
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+
}
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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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+
|
|
171
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+
}
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172
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+
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|
173
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+
```
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