CodeForces #209(Div.2) - AB

A. Table

如果值为1的点中，有在边界上的，输出2；否则输出4

代码：

#include <stdio.h>int main() {int n, m;while(scanf("%d%d", &n, &m)!=EOF) {int map[55][55];for(int i = 1; i <= n; i++) {for(int j = 1; j <= m; j++) {scanf("%d", &map[i][j]);}}int flag = 0;for(int i = 1; i <= n; i++) {for(int j = 1; j <= m; j++) {if(map[i][j]) {if(i == 1 || j == 1 || i == n || j == m) flag++;}}}if(!flag) printf("4\n");elseprintf("2\n");}return 0;}

B. Permutation

默认1到2n的数就是1,2,3,……,2*n-2,2*n-1,2*n, 前k对数是2*i+2和2*i+1，后面的全部调换

代码：

#include <stdio.h>#include <math.h>int main() {int n, k, a[100005];while(scanf("%d%d", &n, &k)!=EOF) {for(int i = 0; i < n; i++) {if(i > 0) printf(" ");if(k > 0)printf("%d %d", 2*i+2, 2*i+1);else printf("%d %d", 2*i+1, 2*i+2);k--;}}return 0;}