HDU 5656 CA Loves GCD dp

CA Loves GCD

题目连接：

http://acm.hdu.edu.cn/showproblem.php?pid=5656

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

Hint

题意

给n个数，然后你可以选择若干个数出来，然后求他的gcd

然后现在让你遍历所有的方案，问你所有方案的和是多少

题解：

dp[i][j]表示现在选了i个数，gcd为j的方案数

dp[i][j]->dp[i+1][j],dp[i+1][gcd(a[i+1],j)]

然后不停转移就好了

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1050;
const int mod = 100000007;
int dp[maxn][maxn],a[maxn],n;
long long ans;
int gcd(int a,int b)
{
    return b==0?a:gcd(b,a%b);
}
void solve()
{
    memset(dp,0,sizeof(dp));dp[0][0]=1;ans=0;
    scanf("%d",&n);
    for(int i=1;i<=n;i++) scanf("%d",&a[i]);
    for(int i=0;i<n;i++)
        for(int j=0;j<maxn;j++)if(dp[i][j])
        {
            (dp[i+1][j]+=dp[i][j])%=mod;
            (dp[i+1][gcd(j,a[i+1])]+=dp[i][j])%=mod;
        }
    for(int i=1;i<maxn;i++) (ans+=1ll*i*dp[n][i]%mod)%=mod;
    printf("%d\n",ans);
}
int main()
{
    int t;scanf("%d",&t);
    while(t--)solve();
}