洛谷 P2375 [NOI2014]动物园

题目传送门

解题思路:

其实对于一个sum[i],其值就等于sum[next[i]] + sum[next[next[i]]] + ... + 1,然后我们可以记忆化,然后题目里又有一个限制,就是前后缀不能重合,那我们就找到最大但是不重合的那种情况,然后往小了枚举就好了.最后答案说要取模,所以不要忘了%%%ljx_xhy_yyq(排名无先后).

AC代码:

#include<iostream>
#include<cstdio>
#include<cstring>
#define ljx_xhy_yyq 1000000007

using namespace std;

int n,sum[1000001],kmp[1000001];
long long ans;
string p,l;

int main() {
    scanf("%d",&n);
    while(n--) {
        cin >> p;
        l = " " + p;
        int len = l.length() - 1,j = 0,u = 1;
        ans = 1;
        memset(sum,0,sizeof(sum));
        sum[1] = 1;
        for(int i = 2;i <= len; i++) {
            while(j != 0 && l[i] != l[j+1]) j = kmp[j];
            if(l[i] == l[j+1]) j++; 
            kmp[i] = j;sum[i] = sum[j] + 1;
        }
        j = 0;
        for(int i = 2;i <= len; i++) {
            while(j != 0 && l[i] != l[j+1]) j = kmp[j];
            if(l[i] == l[j+1]) j++;
            while(j * 2 > i) j = kmp[j];
            ans *= (long long)(sum[j]+1);
            ans %= ljx_xhy_yyq;
        }
        printf("%lld
",ans);
    }

    return 0;
}