hdu-5791 Two(dp)

题目链接：

Two

Time Limit: 2000/1000 MS (Java/Others)    

Memory Limit: 65536/65536 K (Java/Others)

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

 

题意：

 

问这两个序列中有多少对子序列相同;

 

思路：

 

dp啦,dp[i][j]表示a[1,i]与b[1,j]里面有多少对子序列相同;转移的时候就是看a[i]与b[j]相同不？

相同的话dp[i][j]=dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1]+dp[i-1][j-1]+1;

不同的话dp[i][j]=dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1];

 

AC代码:

 

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************************************************ */ 
 
 
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>
 
using namespace std;
 
#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
 
typedef  long long LL;
 
template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + '0');
    putchar('
');
}
 
const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=1e9;
const int N=1e6+10;
const int maxn=1e3+14;
const double eps=1e-8;

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

int main()
{      
        int n,m;
        while(scanf("%d%d",&n,&m)!=EOF)
        {
            For(i,1,n)read(a[i]);
            For(i,1,m)read(b[i]);
            For(i,1,n)
            {
                For(j,1,m)
                {
                    if(a[i]!=b[j])dp[i][j]=(dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1]+mod)%mod;
                    else dp[i][j]=(dp[i-1][j]+dp[i][j-1]+1)%mod;
                }
            }
            printf("%lld
",dp[n][m]);
        }
        return 0;
}