HDU 5656 CA Loves GCD--DP问题CA Loves GCD

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概述

CA Loves GCD

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 2300 Accepted Submission(s): 770

Problem Description

CA is a fine comrade who loves the party and people; inevitably she loves GCD (greatest common divisor) too.
Now, there are N different numbers. Each time, CA will select several numbers (at least one), and find the GCD of these numbers. In order to have fun, CA will try every selection. After that, she wants to know the sum of all GCDs.
If and only if there is a number exists in a selection, but does not exist in another one, we think these two selections are different from each other.

Input

First line contains

T denoting the number of testcases.
T testcases follow. Each testcase contains a integer in the first time, denoting

N, the number of the numbers CA have. The second line is

N numbers.
We guarantee that all numbers in the test are in the range [1,1000].

1≤T≤50

Output

T lines, each line prints the sum of GCDs mod

100000007.

Sample Input

   2 2 2 4 3 1 2 3 
 

Sample Output

   8 10 
 

题意，输入N个数，每一次随机拿出若干个数求他们的gcd并且加在一起，最后输出和是多少。

DP问题，我们设dp【i】【j】中的i表示前i个数字，j表示从前i个数字取出若干个数使他们的gcd是j。

之后会发现每一次求gcd会T，所以需要提前打表计算gcd。

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
long long  a[1005],dp[1005][1005],GCD[1005][1005];
long long  gcd(long long a,long long b)
{
    return b?gcd(b,a%b):a;
}
int main()
{
    for(int i=1; i<=1000; i++)
        for(int j=1; j<=1000; j++)
            GCD[i][j]=gcd(i,j); ///提前打表，否则每一次调用gcd会超时
    int testnum;
    scanf("%d",&testnum);
    while(testnum--)
    {
        int num;
        scanf("%d",&num);
        memset(dp,0,sizeof(dp));
        for(int i=1; i<=num; i++)
            scanf("%d",&a[i]);
        for(int i=1; i<=num; i++)
        {
            dp[i][a[i]]=1; ///表示前i-1个都不取，仅仅取第i个
            for(int j=1; j<=1000; j++)   ///j遍历
            {
                dp[i][j]=(dp[i][j]+dp[i-1][j])%100000007; ///这个数字不取
                if(dp[i-1][j])  ///取这个数字，并求这个数和这个数之前的最大公约数的gcd
                    dp[i][GCD[a[i]][j]]=(dp[i][GCD[a[i]][j]]+dp[i-1][j])%100000007;
                    ///该处不能写成dp[i][GCD[a[i]][j]]=dp[i-1][j]%100000007
                    ///因为本身也要加上，而且GCD[a[i]][j]]一定是比j小的，所以不会漏掉数据
            }
        }
        long long sum=0;
        for(int j=1; j<=1000; j++) ///只需遍历最后，因为之前的方案数都推到后面了
            if(dp[num][j])
                sum=(sum+dp[num][j]*j)%100000007;
        printf("%lldn",sum);
    }
}