hdu 5791 Two dp

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5791

题意：给你两个序列，求相同的子序列个数，结果对1e9+7取余

题解：dp[i][j]表示两序列分别以i,j为结尾的子问题结果，当i不等于j时，d[i][j] = d[i-1][j]+dp[i][j-1]-dp[i-1][j-1]，当i等于j时，d[i][j] = d[i-1][j]+d[i][j-1]+1;

推导过程：当i等于j时，d[i][j] = （i不等于j时的结果） + dp[i-1][j-1] + 1.

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;

const int mod = 1e9 + 7;
const int maxn = 1005;

int dp[maxn][maxn];
int a[maxn], b[maxn];

int main(){
   int n,m;
   while(scanf("%d%d",&n,&m) == 2){
       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]) % mod - dp[i-1][j-1]) % mod;
             else{
                 dp[i][j] = ((dp[i-1][j] + dp[i][j-1]) % mod + 1) % mod;
             }
          }
       }
       printf("%d
",(dp[n][m] + mod) % mod);
   }
   return 0;
}