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时间:2016-08-07 09:38:56      阅读:255      评论:0      收藏:0      [点我收藏+]

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Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1533    Accepted Submission(s): 676

 

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:

求公共子序列数量。

 

dp[i][j]表示第一个串考虑到i位,第二个串考虑到j位的答案是多少。

那么dp[i][j] = dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1] ,需要特别判断a[i]=b[j]时,dp[i][j]+=dp[i-1][j-1]+1。

 

附AC代码:

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 
 4 const int pr=1000000007;
 5 
 6 int dp[2010][2010];
 7 int a[2010],b[2010];
 8 
 9 int main(){
10     int n,m;
11     while(cin>>n>>m){
12         for(int i=1;i<=n;i++){
13             for(int j=1;j<=m;j++){
14                 dp[i][j]=0;
15             }
16         }
17         for(int i=1;i<=n;i++){
18             cin>>a[i];
19         }
20         for(int i=1;i<=m;i++){
21             cin>>b[i];
22         }
23         for(int i=1;i<=n;i++){
24             for(int j=1;j<=m;j++){
25                 dp[i][j]=dp[i][j-1]+dp[i-1][j]-dp[i-1][j-1];
26                 if(a[i]==b[j])
27                 dp[i][j]+=dp[i-1][j-1]+1;
28                 if(dp[i][j]<0)
29                 dp[i][j]+=pr;
30                 if(dp[i][j]>=pr)
31                 dp[i][j]%=pr;
32             }
33         }
34         cout<<dp[n][m]<<endl;
35     }
36     return 0;
37 }

 

Two

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原文地址:http://www.cnblogs.com/Kiven5197/p/5745529.html

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