hdu 5791 Two (dp)

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)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

dp[i][j] 代表第一个串到i，第二个串到j的公共子串和，那么可以得到的是
dp[i][j]=dp[i-1][j]+dp[i][j-1]-dp[i-1]j-1
就是减掉了重复的部分，
如果a[i]!=a[j]
那么 dp[i][j]=dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1]+dp[i-1][j-1]+1;
因为要加上 还没包含i和 j的情况下的所有子序列都可以到达 dp[i][j]，再加上自身贡献+1

#include <bits/stdc++.h>using namespace std;const int mod = 1e9+7;typedef long long ll;int a[1100],b[1100];ll dp[1100][1100];int main(){    int n,m;    while(~scanf("%d%d",&n,&m))    {        memset(dp,0,sizeof(dp));        for(int i=1;i<=n;i++)            scanf("%d",&a[i]);        for(int i=1;i<=m;i++)            scanf("%d",&b[i]);        for(int i=1;i<=n;i++)        {            for(int j=1;j<=m;j++)            {                    if(a[i]==b[j]) dp[i][j]=(dp[i-1][j]+dp[i][j-1]+1),dp[i][j]%=mod;                    else dp[i][j]=(dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1]+mod)%mod;            }        }        printf("%lld\n",dp[n][m] );    }}