HDOJ 5791 Two(dp—求两个序列的公共序列个数)

Two

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 356    Accepted Submission(s): 140

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 21 2 32 13 21 2 31 2

 

Sample Output

23

 

题意：给出A ，B两个序列，求A，B的公共序列个数。

题解：简单dp，板子题，弱逼的我找了板子才A掉。蔡茹苟

代码如下：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define maxn 1010int a[maxn],b[maxn];long long dp[maxn][maxn];const int mod=1000000007;int main(){int n,m,i,j;while(scanf("%d%d",&n,&m)!=EOF){for(i=1;i<=n;++i)scanf("%d",&a[i]);for(i=1;i<=m;++i)scanf("%d",&b[i]);dp[0][0]=dp[1][0]=dp[0][1]=0;memset(dp,0,sizeof(dp));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]-dp[i-1][j-1]+mod)%mod;//这个地方可能会出现负数，要加mod elsedp[i][j]=(dp[i-1][j]+dp[i][j-1]+1)%mod;}}printf("%I64d\n",dp[n][m]);}return 0;}