- 猜谜游戏
- 2016-09-20 21:31:54 @
N^2次循环,最多40000.
2 条评论
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呼噜娃 LV 7 @ 2016-10-08 22:02:20
何人能看懂?
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2016-10-08 22:02:03@
#include <stdio.h> int gg[101]; int g[101]; int s; int a[201][3]; int main() { int c, d; scanf("%d", &c); for ( int e = 0; e < c; e ++){ for ( int f = 0; f < 3; f++) scanf("%d", &a[e][f]); } for ( int f = 0; f < 3; f ++){ for (int e = 0; e < c; e ++){ g[a[e][f]] ++; if (g[a[e][f]] == 1) gg[a[e][f]] = e; if ( g[a[e][f]] > 1){ a[gg[a[e][f]]][f] = 0; a[e][f] = 0; } } for ( int h = 0; h < 101; h ++) gg[h] = 0; for ( int h = 0; h < 101; h ++) g[h] = 0; } printf("\n"); int y[c]; for ( int e = 0; e < c; e ++) y[e] = 0; for ( int e = 0; e < c; e ++){ for (int f = 0; f < 3; f ++){ printf("%d ", a[e][f]); } printf("\n"); //goto kl; } kl: for ( int e = 0; e < c; e ++){ for ( int f = 0; f < 3; f ++){ y[e] += a[e][f]; } } for ( int e = 0; e < c; e ++) printf("%d\n", y[e]); }
- 1
信息
- ID
- 1941
- 难度
- 3
- 分类
- (无)
- 标签
- 递交数
- 897
- 已通过
- 473
- 通过率
- 53%
- 被复制
- 2
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