2016多校训练Contest5: 1011 Two hdu5791

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

直接dp[i][j]暴表示第一个数列前i项第二个前j项的相同种数

然后暴力n^2转移即可

队友写的代码

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <algorithm>
#include <string.h>
#include <iostream>
#include <set>
#include <map>
#include <vector>
#include <queue>
#include <stack>
using namespace std;
const int INF = 2147483647;
const double PI = acos(-1);

/* ----------------- code ----------------- */

int a[1010],dp[1010][1010],b[1010];
const int mod = 1000000007;
int main(void){

int n,m;
while(scanf("%d%d",&n,&m)!=EOF){
for(int i=1;i<=n;i++){
scanf("%d",&a[i]);
}
for(int i=1;i<=m;i++){
scanf("%d",&b[i]);
}
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]) % mod - dp[i - 1][j - 1] ) % mod + mod) % mod + (a[i] == b[j] ? dp[i - 1][j - 1] + 1 : 0) ) % mod;
}
}
/*
for(int i=0;i<=n;i++){
for(int j=0;j<=m;j++){
printf("%d ",dp[i][j]);
}
printf("\n");
}
*/
printf("%d\n",dp
[m]%mod);
}

return 0;
}