题意：

一串链码  用差分的最小表示法表示

思路：

其实就是最小表示法的模版题…

代码：

#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<cstdlib>
#include<ctime>
#include<cmath>
#include<bitset>
using namespace std;
#define N 600010
#define inf 50000000

int n;
int a[N];

int GetMin() {
    int i, j, k, len = n;
    for (i = 0, j = 1; i < len && j < len;) {
        for (k = 0; k < len && a[i + k] == a[j + k]; k++)
            ;
        if (k >= len)
            break;
        if (a[i + k] < a[j + k])
            j += k + 1;
        else
            i += k + 1;
        if (i == j)
            j++;
    }
    return i;
}

char str[N];
int main() {
    while (~scanf("%s", str)) {
        n = strlen(str);
        for (int i = 0; i < n; i++)
            a[i] = str[i] - '0';
        int t = a[0];
        for (int i = 0; i < n - 1; i++)
            a[i] = (a[i + 1] - a[i] + 8) % 8;
        a[n - 1] = (t - a[n - 1] + 8) % 8;
        for (int i = 0; i < n; i++)
            a[i + n] = a[i];
        int i = GetMin();
        for (int j = 0; j < n; j++)
            printf("%d", a[i + j]);
        printf("\n");
    }
    return 0;
}

HDU 4162 Shape Number