• HDU 5464.Clarke and problem-动态规划


    Clarke and problem

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
    Total Submission(s): 1292    Accepted Submission(s): 532


    Problem Description
    Clarke is a patient with multiple personality disorder. One day, Clarke turned into a student and read a book.
    Suddenly, a difficult problem appears: 
    You are given a sequence of number a1,a2,...,an and a number p. Count the number of the way to choose some of number(choose none of them is also a solution) from the sequence that sum of the numbers is a multiple of p(0 is also count as a multiple of p). Since the answer is very large, you only need to output the answer modulo 109+7
     
    Input
    The first line contains one integer T(1T10) - the number of test cases. 
    T test cases follow. 
    The first line contains two positive integers n,p(1n,p1000) 
    The second line contains n integers a1,a2,...an(|ai|109).
     
    Output
    For each testcase print a integer, the answer.
     
    Sample Input
    1 2 3 1 2
     
    Sample Output
    2 Hint: 2 choice: choose none and choose all.
     
    Source

    代码:

     1 #include<stdio.h>
     2 #include<string.h>
     3 const int mod=1e9+7;
     4 const int N=1000+10;
     5 int a[N];
     6 int dp[N][N];
     7 int main(){
     8     int t,n,p;
     9     while(~scanf("%d",&t)){
    10         while(t--){
    11             scanf("%d%d",&n,&p);
    12             for(int i=1;i<=n;i++){
    13                 scanf("%d",&a[i]);
    14                 a[i]=(a[i]%p+p)%p;
    15             }
    16             dp[0][0]=1;
    17             for(int i=1;i<=n;i++){
    18                 for(int j=0;j<=p-1;j++){
    19                     dp[i][j]=(dp[i-1][j]+dp[i-1][(j-a[i]+p)%p])%mod;
    20                 }
    21             }
    22             printf("%d
    ",dp[n][0]%mod);
    23         }
    24     }
    25     return 0;
    26 }
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  • 原文地址:https://www.cnblogs.com/ZERO-/p/9741072.html
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