題解:
dp[i][j]表示A序列前i個數和B序列前j個數的相同子序列對有多少個。複雜度O(n^2)O(n2)
看了題解就會了。。。。。
ACcode:
#include <bits/stdc++.h>
#define maxn 1000
#define mod 1000000007
#define ll long long
using namespace std;
int a[maxn],b[maxn];
ll dp[maxn][maxn];
int main(){
int n,m;
while(scanf("%d%d",&n,&m)!=EOF){
for(int i=1;i<=n;++i)scanf("%d",&a[i]);
for(int j=1;j<=m;++j)scanf("%d",&b[j]);
memset(dp,0,sizeof(dp));
for(int i=1;i<=n;++i)
for(int j=1;j<=m;++j){
dp[i][j]=(dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1]+mod)%mod;
if(a[i]==b[j])
dp[i][j]=(dp[i-1][j]+dp[i][j-1]+1+mod)%mod;
}
cout<<dp[n][m]<<'\12';
}
return 0;
}