POJ

传送门

最小表示法模板

//Achen
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<vector>
#include<queue>
#include<cmath>
#include<ctime>
const int N=100007;
typedef long long LL;
using namespace std;
int T,n;
char s[N]; 

template<typename T> void read(T &x) {
    T f=1; x=0; char ch=getchar();
    while(ch!='-'&&(ch<'0'||ch>'9')) ch=getchar();
    if(ch=='-') f=-1,ch=getchar();
    for(;ch>='0'&&ch<='9';ch=getchar()) x=x*10+ch-'0'; x*=f;
}

int solve() {
    int i=0,j=1,k=0;
    for(;i<n&&j<n&&k<n;) {
        int t=s[(i+k)%n]-s[(j+k)%n];
        if(!t) k++;
        else {
            if(t>0) i=i+k+1; 
            else j=j+k+1;
            if(i==j) j++;
            k=0;
        }
    } 
    return min(i,j);
}

int main() {
    read(T);
    while(T--) {
        scanf("%s",s);
        n=strlen(s);
        printf("%d
",solve()+1);
    }
    return 0;
}