hdu5791

Problem Description

Alice gets two sequences A and B. A easy problem comes. How many pair of sequence A' and sequence B' are same. For example, {1,2} and {1,2} are same. {1,2,4} and {1,4,2} are not same. A' is a subsequence of A. B' is a subsequence of B. The subsequnce can be not continuous. For example, {1,1,2} has 7 subsequences {1},{1},{2},{1,1},{1,2},{1,2},{1,1,2}. The answer can be very large. Output the answer mod 1000000007.

 

Input

The input contains multiple test cases.

For each test case, the first line cantains two integers N,M(1≤N,M≤1000). The next line contains N integers. The next line followed M integers. All integers are between 1 and 1000.

 

Output

For each test case, output the answer mod 1000000007.

这题的大意是给你两个序列，让你求两个序列的公共子序列有多少个。

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;long long dp[1100][1100];const int mod = 1000000007;int main(void){    int n,m,i,j;    int a[1100],b[1100];    while(scanf("%d%d",&n,&m)==2)    {        for(i=1;i<=n;i++)            scanf("%d",&a[i]);        for(i=1;i<=m;i++)            scanf("%d",&b[i]);        memset(dp,0,sizeof(dp));        for(i=1;i<=n;i++)            for(j=1;j<=m;j++)            {                dp[i][j] = ((dp[i-1][j] + dp[i][j-1])%mod - dp[i-1][j-1] + mod)%mod;                dp[i][j] %= mod;                if(a[i] == b[j])                    dp[i][j] = (dp[i][j]+ dp[i-1][j-1])%mod + 1;                dp[i][j] %= mod;            }        cout << dp[n][m] << endl;    }}