hdu 5791 (DP) Two

小鱼儿 2021-11-10 08:44 148阅读 0赞

[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(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

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 }

转载于:https://www.cnblogs.com/JJCHEHEDA/p/5738830.html

[hdu 5791]: http://acm.hdu.edu.cn/showproblem.php?pid=5791