hdu 5791 Two 二维前缀和

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.

才发现这道题是多校的。。。感觉还是比较水的。
公式比较好推，我用dp[i][j]表示以a的第i位和b的第j位结尾的符合题意的子串的个数，sum【i】【j】则是dp的一个二维前缀和。二维前缀和怎么求呢？可以百度二维前缀和23333.
然后假如a【i】==b【j】那么dp【i】【j】=sum【i-1】【j-1】，否则为0，算出dp后更新sum的值。
另外还有一个坑点，WA了几发，就是出现负数要还原，这点在有mod的题目里很重要。

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<limits.h>
#include<float.h>
#include<algorithm>
using namespace std;
const int MOD=1000000007;
int dp[1010][1010],sum[1010][1010],a[1010],b[1010],n,m;
int main(void)
{
while(cin>>n>>m)
{
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
for(int i=1;i<=m;i++)
scanf("%d",&b[i]);
memset(dp,0,sizeof(dp));
memset(sum,0,sizeof(sum));
for(int i=1;i<=n;i++)
{
for(int j=1;j<=m;j++)
{
sum[i][j]=(sum[i-1][j]+sum[i][j-1]-sum[i-1][j-1])%MOD;
sum[i][j]=(sum[i][j]+MOD)%MOD;
if(a[i]==b[j])
dp[i][j]=(sum[i-1][j-1]+1)%MOD;
else
dp[i][j]=0;
sum[i][j]=(sum[i][j]+dp[i][j])%MOD;
}
}
cout<<sum
[m]<<endl;

}
return 0;
}