【BZOJ 1398】 Necklace

【题目链接】

https://www.lydsy.com/JudgeOnline/problem.php?id=1398

【算法】

最小表示法

【代码】

#include<bits/stdc++.h>
using namespace std;
#define MAXN 1000010

char s1[MAXN<<],s2[MAXN<<],a[MAXN],b[MAXN];

inline void get(char *t,char *a)
{
        int i,j,k,n,pos;
        n = strlen(a+);
        for (i = ; i <= n; i++) a[i+n] = a[i];
        i = ; j = ;
        while (i <= n && j <= n)
        {
                for (k = ; k <= n && a[i+k] == a[j+k]; k++);
                if (k == n) break;
                if (a[i+k] > a[j+k])
                {
                        i = i + k + ;
                        if (i == j) i++;
                }    else
                {
                        j = j + k + ;
                        if (i == j) j++;
                }
        }
        pos = min(i,j);
        for (i = pos; i <= n; i++) t[i-pos] = a[i];
        for (i = ; i < pos; i++) t[n-pos+i] = a[i];
}

int main()
{

        scanf("%s%s",s1+,s2+);
        get(a,s1);
        get(b,s2);
        if (strcmp(a,b) == ) printf("Yes\n%s\n",a);
        else printf("No\n");

        return ;

}