回答編集履歴
2
ビット演算一般形
test
CHANGED
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@@ -59,3 +59,81 @@
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59
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}
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```
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62
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+
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63
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Javaの方がなじみあるのでJavaで書いてみますね。
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64
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65
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BigInteger使うと簡単なんだろうけど、C++に標準の多倍長整数型がないようなので、配列で頑張った
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66
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+
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67
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最後だけめんどくさかったのでBitSetに変換して出力
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68
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69
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```java
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70
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import java.util.*;
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72
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74
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public class Main {
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public static void main(String[] args) {
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78
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79
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try (Scanner s = new Scanner(System.in)) {
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80
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81
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int n = s.nextInt();
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82
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int sum = 0;
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84
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85
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int[] a = new int[n];
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86
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87
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for (int i = 0; i < n; i++) {
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a[i] = s.nextInt();
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90
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91
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sum += a[i];
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92
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93
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}
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94
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95
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long[] bit = new long[(sum >> 6) + 1];
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96
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97
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bit[0] = 1;
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98
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sum = 0;
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100
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103
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for (int i = 0; i < n; i++) {
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int shiftCell = a[i] >> 6;
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int shiftBit = a[i] & 0x3f;
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108
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109
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for (int j = (sum >> 6); j >= 0; j--) {
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110
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111
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if (shiftBit != 0) {
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112
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long upper = bit[j] >>> (64 - shiftBit);
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114
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115
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if (upper != 0) {
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bit[j + shiftCell + 1] |= upper;
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}
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120
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}
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122
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bit[j + shiftCell] |= bit[j] << shiftBit;
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125
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}
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126
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127
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sum += a[i];
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128
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129
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}
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130
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System.out.println(BitSet.valueOf(bit));
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133
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}
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134
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135
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}
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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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```
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1
=が必要だった
test
CHANGED
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@@ -40,7 +40,7 @@
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40
40
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43
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-
for(int S = 0; S < max; S++) {
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for(int S = 0; S <= max; S++) {
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if(x & (1 << S)) {
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