hdu 5791

Two

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1421    Accepted Submission(s): 630


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(1N,M1000). 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
 

求两个序列的相同的公共子序列的个数,取模。

DP[i][j]表示到a数组的第 i 个和b数组的第 j 个之间的个数,可以归纳出,当a[i]==b[j]时,dp[i][j]=dp[i-1][j]+dp[i][j-1]+1;

当不等的时候,dp[i][j]=dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1].(仔细琢磨琢磨,会发现很有道理)因为有减法的运算所以最后答案可能出现

负数,注意处理一下。

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<iostream>
 4 #include<algorithm>
 5 using namespace std;
 6 
 7 typedef long long ll;
 8 const int mod = 1000000007;
 9 ll dp[1010][1010];
10 int a[1010],b[1010];
11 
12 int main()
13 {
14     int n,m;
15     while (~scanf("%d%d",&n,&m)){
16         for (int i=1 ; i<=n ; i++) scanf("%d",&a[i]);
17         for (int i=1 ; i<=m ; i++) scanf("%d",&b[i]);
18         memset(dp,0,sizeof(dp));
19         for (int i=1 ; i<=n ; i++){
20             for (int j=1 ; j<=m ; j++){
21                 if (a[i]==b[j]) dp[i][j]=dp[i-1][j]+dp[i][j-1]+1;
22                 else dp[i][j]=dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1];
23                 dp[i][j]%=mod;
24             }
25         }
26         printf("%I64d\n",(dp[n][m]+mod)%mod);
27     }
28     return 0;
29 }

 

 
posted on 2016-08-04 23:22  蜘蛛侦探  阅读(221)  评论(0编辑  收藏  举报