wangli2029的博客

/* What Are You Talking About Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 102400/204800 K (Java/Others) Total Submission(s): 19289 Accepted Submission(s): 6313 Problem Description I

/* 统计难题 Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131070/65535 K (Java/Others) Total Submission(s): 29864 Accepted Submission(s): 11652 Problem Description Ignatius最近遇到一个难题,老师交给他很多单

题目链接  http://poj.org/problem?id=2752 给定一个字符串s，从小到大输出s中既是前缀又是后缀的子串的长度。 #include #include char str[400005]; int next[400005]; int sum[400005]; int len; int count; void getnext( ) { int i,j; next

题目链接：http://poj.org/problem?id=1961 题目大意： 给你一个字符串，求这个字符串到第i个字符为还是用KMP算法 算是2406的加强版  不过从头到尾循环遍历一次就行了 #include char str[1000005]; int next[1000005]; int t=0; int len; void getnext() { int i=0,j=-1;

题目链接：http://poj.org/problem?id=2406/*#include int main() { printf("%d\n",0%4);//被求余数不能为0 return 0; } */ #include #include char s[1000005]; int next[1000005]; int len; void getnext() { int i=0,j