2015 HUAS Summer Contest#4~F

Description

A subsequence of a given sequence is the given sequence with some elements (possible none) left out. Given a sequence X = < x1, x2, ..., xm > another sequence Z = < z1, z2, ..., zk > is a subsequence of X if there exists a strictly increasing sequence < i1, i2, ..., ik > of indices of X such that for all j = 1,2,...,k, x _ij = zj. For example, Z = < a, b, f, c > is a subsequence of X = < a, b, c, f, b, c > with index sequence < 1, 2, 4, 6 >. Given two sequences X and Y the problem is to find the length of the maximum-length common subsequence of X and Y.

Input

The program input is from the std input. Each data set in the input contains two strings representing the given sequences. The sequences are separated by any number of white spaces. The input data are correct.

Output

For each set of data the program prints on the standard output the length of the maximum-length common subsequence from the beginning of a separate line.

Sample Input

abcfbc         abfcab
programming    contest 
abcd           mnp

Sample Output

4
2
0
解题思路：题目的意思是求两个字符串的最长子序列。但是需要注意的是：子序列不是子串，子序列这些字符可以是不连续的  但是有先后顺序；而子串是连续的。

程序代码：

#include<stdio.h>
#include<string.h>
const int maxn=1005;
char a[maxn],b[maxn];
int d[maxn][maxn];
int max(int x,int y)
{
    if(x>y) return x;
    else return y;
}
int main()
{
    int i,j,t,floag,n,m;
    while(scanf("%s%s",a+1,b+1)!=EOF)
    {
        memset(d,0,sizeof(d));
        n=strlen(a+1);
        m=strlen(b+1);
        for(i=1;i<=n;i++)
        for(j=1;j<=m;j++)
        if(a[i]==b[j])  d[i][j]=d[i-1][j-1]+1;
        else  d[i][j]=max(d[i-1][j],d[i][j-1]);
        printf("%d\n",d[n][m]);
    }
    return 0;
}