HDU 5464 Clarke and problem(dp 动态规划)——BestCoder Round #56(div.1 div.2)
Clarke and problem
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
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
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(1≤T≤10) - the number of test cases.
T test cases follow.
The first line contains two positive integers n,p(1≤n,p≤1000)
The second line contains n integers a1,a2,...an(|ai|≤109 ).
T test cases follow.
The first line contains two positive integers n,p(1≤n,p≤1000)
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
BestCoder Round #56 (div.2)
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附上该题对应的中文题
Clarke and problem
Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 65536/65536 K (Java/Others)
问题描述
克拉克是一名人格分裂患者。某一天，克拉克分裂成了一个学生，在做题。 突然一道难题难到了克拉克，这道题是这样的： 给你n个数，要求选一些数（可以不选），把它们加起来，使得和恰好是p的倍数（0也是p的倍数），求方案数。 对于n很小的时候，克拉克是能轻易找到的。然而对于n很大的时候，克拉克没有办法了，所以来求助于你。
输入描述
第一行一个整数T(1≤T≤10)，表示数据的组数。 每组数据第一行是两个正整数n,p(1≤n,p≤1000)。 接下来的一行有n个整数ai(∣ai∣≤109)，表示第i个数。
输出描述
对于每组数据，输出一个整数，表示问题的方案数，由于答案很大，所以求出对109+7的答案即可。
输入样例
1 2 3 1 2
输出样例
2
Hint
有两种方案：什么也不选；全都选。
/****************************************************/
好吧，那只能转换思路了，要使相加之和的结果是p的倍数，就是说%p==0，那么我们不妨一开始就将ai处理成<p的数，以便用数组进行操作
剩下的工作就是用dp解决问题了，转移方程是前i个数，模p为j的方案数dp[i][j]=dp[i-1][j]+dp[i-1][(j-a[i]+p)%p]
即第i个数不取的情况，和取了第i个数的情况。
①若第i个数不取，那么dp[i][j]=dp[i-1][j]，不影响方案数；
②若取了第i个数，就相当于前i-1个数相加模p的值为dp[i-1][(j-a[i]+p)%p]，即减去第i项值a[i]
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<queue>
#include<stack>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<stdlib.h>
#include<cmath>
#include<string>
#include<algorithm>
#include<iostream>
#define exp 1e-10
using namespace std;
const int N = 1001;
const int inf = 1000000000;
const int mod = 1000000007;
__int64 s[N][N];
int a[N];
int main()
{
int t,n,p,i,j;
scanf("%d",&t);
while(t--)
{
memset(s,0,sizeof(s));
scanf("%d%d",&n,&p);
for(i=1;i<=n;i++)
{
scanf("%d",&a[i]);
a[i]%=p;
}
for(s[0][0]=1,i=1;i<=n;i++)
for(j=0;j<=p;j++)
s[i][j]=(s[i-1][j]+s[i-1][(j-a[i]+p)%p])%mod;
printf("%I64d\n",s[n][0]);
}
return 0;
}
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