链接：
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 1462 Accepted Submission(s): 521
Problem Description
Today is the 1st anniversary of BestCoder. Soda, the contest manager, gets a string s of length n. He wants to find three nonoverlapping substrings s[l1..r1], s[l2..r2], s[l3..r3] that:1. 1≤l1≤r1<l2≤r2<l3≤r3≤n2. The concatenation of s[l1..r1], s[l2..r2], s[l3..r3] is "anniversary".
Input
There are multiple test cases. The first line of input contains an integer T (1≤T≤100), indicating the number of test cases. For each test case:There's a line containing a string s (1≤|s|≤100) consisting of lowercase English letters.
Output
For each test case, output "YES" (without the quotes) if Soda can find such thress substrings, otherwise output "NO" (without the quotes).
Sample Input
2 annivddfdersewwefary nniversarya
Sample Output
YES NO
代码：
#include#include #include #define N 150char s1[20] = "anniversary";char ch[N];char s[N][3][20];void Init(){ int j, z, w=0; for(j=1; j<=9; j++) for(z=1; z<=10-j; z++) { strncpy(s[w][0], s1, j); strncpy(s[w][1], s1+j, z); strcpy(s[w++][2], s1+j+z); }}int main(){ int t; scanf("%d", &t); Init(); while(t--) { char ch[N]; int i, a, b, len1, len2, flag=0; memset(ch, 0, sizeof(ch)); scanf("%s", ch); for(i=0; i<45; i++) { if(strstr(ch, s[i][0])) { a = strstr(ch, s[i][0])-ch; len1 = strlen(s[i][0]); a = a + len1; if(strstr(ch+a, s[i][1])) { b = strstr(ch+a, s[i][1])-(ch+a); ///就在这， 我在判断的时候还把字符串向后移了 a 位， 但在 ///计算的时候并没有用，这是最大的失误，难怪自己思路对了但一直提交错误 len2 = strlen(s[i][1]); b = a + b + len2; if(strstr(ch+b, s[i][2])) { flag = 1; break; } } } } if(flag) printf("YES\n"); else printf("NO\n"); } return 0;}