考慮dp[i]代表前綴s[1...i]出現的次數，必定有dp[nxt[i]] += dp[i]
倒著推就是了

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
int T, n, nxt[200005], dp[200005], ans;
const int mod=10007;
char a[200005];
void mknxt(){
    int k=0;
    for(int i=2; i<=n; i++){
        while(k && a[i]!=a[k+1
 
])    k = nxt[k];
        if(a[i]==a[k+1])    nxt[i] = ++k;
    }
}
int main(){
    cin>>T;
    while(T--){
        ans = 0;
        scanf("%d", &n);
        scanf("%s", a+1);
        memset(nxt, 0, sizeof(nxt));
        for(int i=1; i<=n; i++)    dp[i] = 1;
        mknxt();
        for(int
 
 i=n; i>=1; i--)
            dp[nxt[i]] = (dp[nxt[i]] + dp[i]) % mod;
        for(int i=1; i<=n; i++)
            ans = (ans + dp[i]) % mod;
        printf("%d\n", ans);
    }
    return 0;
}

hdu3366 Count the string