hdu 5791 Two dp

Two

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

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.

 

Sample Input

3 2 1 2 3 2 1 3 2 1 2 3 1 2

 

Sample Output

2 3

 

Author

ZSTU

 

Source

2016 Multi-University Training Contest 5

思路：if(a[i]==b[t])
         dp[i][t]=((1+dp[i][t-1]+dp[i-1][t])%mod+mod)%mod;
         else
         dp[i][t]=((dp[i][t-1]+dp[i-1][t]-dp[i-1][t-1])%mod+mod)%mod;

　　  dp[i][t]表示答案，A数组的i位置，b数组的t位置；

　　  利用前缀减少时间复杂度，从前面继承；

　　  ps：又忘了取模的坑点，减法需要+mod%mod；

#include<bits/stdc++.h>
using namespace std;
#define ll __int64
#define esp 0.00000000001
const int N=1e3+10,M=1e7+10,inf=1e9+10,mod=1000000007;
int a[N];
int b[N];
ll dp[N][N];
int main()
{
    int x,y,z,i,t;
    while(~scanf("%d%d",&x,&y))
    {
        memset(dp,0,sizeof(dp));
        for(i=1;i<=x;i++)
        scanf("%d",&a[i]);
        for(i=1;i<=y;i++)
        scanf("%d",&b[i]);
        for(i=1;i<=x;i++)
        {
            for(t=1;t<=y;t++)
            if(a[i]==b[t])
            dp[i][t]=((1+dp[i][t-1]+dp[i-1][t])%mod+mod)%mod;
            else
            dp[i][t]=((dp[i][t-1]+dp[i-1][t]-dp[i-1][t-1])%mod+mod)%mod;
        }
        printf("%I64d
",dp[x][y]);
    }
    return 0;
}