HUST1010-The Minimum Length

最新推荐文章于 2019-04-03 20:24:10 发布

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本文介绍了一种解决特定字符串问题的方法，即从给定的重复切割字符串中找到最短的原始字符串。通过使用字符串匹配算法，实现了高效计算。

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The Minimum Length

时间限制：1秒内存限制：128兆

2266 次提交 837 次通过

题目描述

There is a string A. The length of A is less than 1,000,000. I rewrite it again and again. Then I got a new string: AAAAAA...... Now I cut it from two different position and get a new string B. Then, give you the string B, can you tell me the length of the shortest possible string A. For example, A="abcdefg". I got abcdefgabcdefgabcdefgabcdefg.... Then I cut the red part: efgabcdefgabcde as string B. From B, you should find out the shortest A.

输入

Multiply Test Cases. For each line there is a string B which contains only lowercase and uppercase charactors. The length of B is no more than 1,000,000.

输出

For each line, output an integer, as described above.

样例输入

bcabcab
efgabcdefgabcde

样例输出

3
7

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>

using namespace std;

#define N 1000005
char str[N];
int f[N];

int main()
{
    while(~scanf("%s",str))
    {
        int len=strlen(str);
        f[0]=-1;
        for(int i=0;i<len;i++)
        {
            int k=f[i];
            while(k>=0&&str[i]!=str[k]) k=f[k];
            f[i+1]=k+1;
        }
        printf("%d\n",len-f[len]);
    }
    return 0;
}