CodeForces 314 B.Sereja and Periods 思维+简单dp【转】

最新推荐文章于 2020-04-25 07:59:31 发布

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最新推荐文章于 2020-04-25 07:59:31 发布

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分类专栏： ACM DP 文章标签： dp 暑假集训

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本文通过一道具体的字符串匹配题目，介绍了如何使用动态规划方法解决最大重复子串的问题。文章详细解释了状态定义、状态转移方程及其实现过程。

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Description

Let's introduce the designation $\text{[math]}$ , where x is a string, n is a positive integer and operation " + " is the string concatenation operation. For example, [abc, 2] = abcabc.

We'll say that string scan be obtained from string t, if we can remove some characters from string t and obtain string s. For example, strings ab and aсba can be obtained from string xacbac, and strings bx and aaa cannot be obtained from it.

Sereja has two strings, w = [a, b] and q = [c, d]. He wants to find such maximum integer p(p > 0), that [q, p] can be obtained from string w.

Input

The first line contains two integers b, d(1 ≤ b, d ≤ 10⁷). The second line contains string a. The third line contains string c. The given strings are not empty and consist of lowercase English letters. Their lengths do not exceed 100.

Output

In a single line print an integer — the largest number p. If the required value of p doesn't exist, print 0.

Sample Input

Input

10 3
abab
bab

Output

题解：

自己用模拟做了n遍都没过。。惭愧，看了博客才恍然大悟要用dp，附上神犇博客http://blog.youkuaiyun.com/yskyskyer123/article/details/52352107 ，因为懒得讲解了，到博客里看吧

引用：

令dp[x]表示c串的第x字符开始匹配，匹配完一个字符串a后，能够匹配多少个整字符串c，并且nex[x]代表下次开始匹配的位置。

代码：

#include<stdio.h>
#include<cstring>
#include<string>
#include<iostream>
#include<algorithm>
#include<math.h>
#include<queue>
#include<stack>
using namespace std;
char s1[1005],s2[1005];
int dp[105];
int nex[105];
int b,d,len1,len2;
void cope(int x)
{
    dp[x]=0;
    int use=0;
    nex[x]=x;
    while(use<len1)
    {
        if(s1[use]==s2[nex[x]])
        {
            nex[x]++;
            if(nex[x]==len2)
            {
                dp[x]++;
                nex[x]=0;
            }
        }
        use++;
    }
}
int main()
{
    int i,j;
    while(scanf("%d%d",&b,&d)!=EOF)
    {
        scanf("%s%s",s1,s2);
        len1=strlen(s1);
        len2=strlen(s2);
        for(i=0;i<len2;i++)
            cope(i);
        int x=0,num=0;
        for(i=0;i<b;i++)
        {
            num+=dp[x];
            x=nex[x];
        }
        num/=d;
        printf("%d\n",num);
    }
    return 0;
}