hdu 5791 Two(LCS)

dp[i][j]表示A序列前i个数和B序列前j个数的相同子序列对有多少个。复杂度O(n^2)O(n?2??)

/*by*/
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
using namespace std;
typedef long long LL;
const LL N=1010;
const LL mod=1000000007;
LL a[N],b[N];
LL dp[N][N];
int main()
{
    LL n,m,i,j;
    while(~scanf("%lld%lld",&n,&m)) {
        memset(dp,0,sizeof(dp));
        for(i=1; i<=n; i++)
            scanf("%lld",&a[i]);
        for(i=1; i<=m; i++)
            scanf("%lld",&b[i]);
        for(i=1; i<=n; i++) {
            for(j=1; j<=m; j++) {
                if(a[i]==b[j])
                    dp[i][j]=(dp[i-1][j]+dp[i][j-1]+1)%mod;
                else
                    dp[i][j]=(dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1]+mod)%mod;/*去重，相减可能是负数，先加个mod*/
            }
        }
        printf("%lld\n",dp[n][m]);
    }
    return 0;
}