hdu 5791 Two 二维dp

Two

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 889    Accepted Submission(s): 405

Problem Description

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.

 

Sample Input

3 2 1 2 3 2 1 3 2 1 2 3 1 2

 

Sample Output

2 3

 

Author

ZSTU

 

Source

2016 Multi-University Training Contest 5

 题意:给你a,b两个数组，可以从a数组中按顺序选出一些数字，然后b数组按顺序选出一些数字，要求

最后选出的数字按顺序相同，有多少选法。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <cstring>
#include <string>
#include <algorithm>
using namespace std;
typedef  long long  ll;
typedef unsigned long long ull;
#define MM(a,b) memset(a,b,sizeof(a));
#define inf 0x7f7f7f7f
#define FOR(i,n) for(int i=1;i<=n;i++)
#define CT continue;
#define PF printf
#define SC scanf
const int mod=1000000007;
const int N=1000+10;
ull seed=13331;
ll dp[N][N];
int a[N],b[N];

int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m))
    {
        for(int i=1;i<=n;i++) scanf("%d",&a[i]);
        for(int i=1;i<=m;i++) scanf("%d",&b[i]);

        MM(dp,0);
        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][j]+dp[i-1][j-1]+1)%mod;
                   }
        printf("%lld
",dp[n][m]);
    }
    return 0;
}

　　分析：昨天比赛的时候一直想着怎么把暴力的复杂度降下去，想到了哈希，大数。。。就是将各种情况的对应到一个数值，，，，然并卵，，因为这道题目取的数字可以是不连续的，哈希跟大数就跪了。

正确解法：设dp[i][j]为当前枚举到a是i位,b是j位的时候，那么先撇开（i,j）不看，那么

dp[i][j]=dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1]；（注意dp[i-1][j-1]多算了一次）

如果a[i]==b[j]，那么（i,j）还可以单独的产生组合，；

最后要注意%mod时，两者相减可能会产生负数

感觉取得数字要是不是连续的话一般就要上dp了