huangrenheluogu @ 2023-10-08 15:57:53
rt。
我是用 P4931 的思路，对于每一个 tem），之后延申到比 tem *= ((i % tem) ? (i / tem) : (i / tem - 1));），然后累加答案。
by huangrenheluogu @ 2023-10-08 15:58:17
代码如下：
#include<bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5;
int n, kmp[N], j, tem;
string s;
long long ans;
int main(){
scanf("%d", &n);
cin >> s;
s = ' ' + s;
for(int i = 2; i <= n; i++){
while(s[i] != s[j + 1] && j) j = kmp[j];
j += (s[i] == s[j + 1]);
kmp[i] = j;
}
for(int i = 1; i <= n; i++){
if(kmp[i] == 0) continue ;
tem = i - kmp[i];
tem *= ((i % tem) ? (i / tem) : (i / tem - 1));
ans += 1ll * tem;
// printf("%d %d\n", i, tem);
}
printf("%lld", ans);
return 0;
}by lightovernight @ 2023-10-12 19:23:03
12 acdacfacdacf 答案应该为48，你这个是42
by huangrenheluogu @ 2023-10-17 20:37:47
@lightovernight 谢谢大神