解题思路:
求前缀数组出现的次数之和,next[i] > 0 表示长度为next[i]的前缀又出现了一次。
#include <iostream> #include <cstring> #include <cstdlib> #include <cstdio> #include <algorithm> #include <cmath> #include <vector> #include <queue> #include <stack> #define LL long long #define FOR(i,x,y) for(int i=x;i<=y;i++) using namespace std; const int maxn = 200000 + 10; const int MOD = 10007; char s[maxn]; int next[maxn]; int main() { int T; scanf("%d", &T); while(T--) { int m; scanf("%d", &m); scanf("%s", s); next[0] = 0; next[1] = 0; for(int i=1;i<m;i++) { int j = next[i]; while(j && s[i] != s[j]) j = next[j]; next[i+1] = (s[i] == s[j]) ? j + 1 : 0; } int ans = m; for(int i=1;i<=m;i++) { if(next[i] > 0) ans = (ans + 1) % MOD; } printf("%d\n", ans); } return 0; }
HDU 3336 Count the string(KMP算法next数组的应用)
原文地址:http://blog.csdn.net/moguxiaozhe/article/details/43446837