hdu6761lyndon分解

题：http://acm.hdu.edu.cn/showproblem.php?pid=6761

分析：每个位置的答案就是加入当前的字符后，该字符串lyndon分解后最后一个lyndon的最左边的位置

lyndon分解参考：https://blog.csdn.net/wayne_lee_lwc/article/details/107528945

#include<bits/stdc++.h>
using namespace std;
#define pb push_back
#define MP make_pair
typedef long long ll;
const int mod=1e9+7;
const int M=2e6+6;
const int inf=0x3f3f3f3f;
const ll INF=1e18;
ll ans[M];
char s[M];


int main(){

    int T;
    scanf("%d",&T);
    while(T--){
        scanf("%s",s);
        int len=strlen(s);
        ans[0]=1;
        int nowi=0;
        while(nowi<len){
            int k=nowi,j=nowi+1;
            while(j<len&&s[j]>=s[k]){
                if(s[j]>s[k]){
                    ans[j]=nowi+1;
                    k=nowi;
                }
                else{
                    ans[j]=ans[k]+j-k;
                    k++;
                }
                j++;
            }
            while(nowi<=k)
                nowi+=j-k;
            if(nowi==j&&nowi<len)
                ans[j]=nowi+1;
        }
        ll res=0ll;
        for(int i=len-1;i>=0;i--)
            res=(res*1112ll%mod+ans[i])%mod;
        printf("%lld\n",res);
    }
    return 0;
}