HDU 3336 Count the string （kmp+dp）

题意:

求串的所有前缀在串中出现次数的总和

分析:

dp[i]:=前i个字符中出现以第i个字符结尾的前缀的次数

dp[i]=dp[nxt[i]]+1

即第i个字符结尾的前缀数等于以第next[i]个字符为结尾的前缀数加上它自己本身,这俩是同一个字符

　 i: 1 2 3 4

　 c: a b a b

dpi: 1 1 2 2

dp数组正好是原来那个倒过来的

代码:

//
//  Created by TaoSama on 2015-10-30
//  Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
const int N = 2e5 + 10, INF = 0x3f3f3f3f, MOD = 1e4 + 7;

int n, nxt
, dp
;
char s
;

int getNxt() {
nxt[0] = -1;
int i = 0, j = -1;
while(i < n) {
if(j == -1 || s[i] == s[j]) nxt[++i] = ++j;
else j = nxt[j];
}
}

void add(int& x, int y) {if((x += y) >= MOD) x %= MOD;}

int main() {
#ifdef LOCAL
freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);

int t; scanf("%d", &t);
while(t--) {
scanf("%d%s", &n, s);
getNxt();
int ans = 0;
for(int i = 1; i <= n; ++i) {
dp[i] = dp[nxt[i]] + 1;
add(ans, dp[i]);
}
printf("%d\n", ans);
}
return 0;
}