UVa 10340 All in All

3-9

本题一遍AC，主要思路是因为字符串序列不变，只要按顺序向后读字符串即可，代码如下：

#include <stdio.h>#include <string.h>char s[100000], t[100000];int main(void){int i, j, n, count = 0;while (1 == scanf("%s", s)) {getchar();scanf("%s", t);getchar();n = strlen(s);s[n] = '\0';n = strlen(t);t[n] = '\0';j = 0;for (i = 0; s[i] != '\0'; i++)for (; t[j] != '\0'; j++)if (s[i] == t[j]) {count++;j++;break;}if (count == strlen(s))printf("Yes\n");elseprintf("No\n");count = 0;}return 0;}