hdu 5656 CA Loves GCD dp

CA Loves GCD

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)

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

 

 

Source

BestCoder Round #78 (div.2)

By YJQ 我们令dp[i][j]表示在前i个数中，选出若干个数使得它们的gcd为j的方案数，于是只需要枚举第i+1个数是否被选中来转移就可以了

令第i+1个数为v，当考虑dp[i][j]的时候，我们令dp[i+1][j] += dp[i][j]dp[i+1][j]+=dp[i][j](v 不选),dp[i+1][gcd(j,v)] += dp[i][j]dp[i+1][gcd(j,v)]+=dp[i][j](v 选)

复杂度O(N*MaxV) MaxV 为出现过的数的最大值

ps：取模的那个数是坑点。。。。平常是1e9+7。。。

#include<bits/stdc++.h>
using namespace std;
#define ll __int64
#define mod 100000007
#define esp 0.00000000001
const int N=1e3+10,M=1e6+10,inf=1e9;
ll dp[N][N];
int a[N];
int flag[N][N];
int gcd(int x,int y)
{
    return y==0?x:gcd(y,x%y);
}
int main()
{
    int x,y,z,i,t;
    for(i=1;i<=1000;i++)
    for(t=1;t<=1000;t++)
    flag[i][t]=gcd(i,t);
    int T;
    scanf("%d",&T);
    while(T--)
    {
        memset(dp,0,sizeof(dp));
        int maxx=0;
        scanf("%d",&x);
        for(i=1;i<=x;i++)
        {
            scanf("%d",&a[i]);
            maxx=max(a[i],maxx);
        }
        for(i=1;i<=x;i++)
        {
            dp[i][a[i]]+=1;
            for(t=1;t<=maxx;t++)
            {
                dp[i][t]+=dp[i-1][t];
                dp[i][t]%=mod;
                dp[i][flag[t][a[i]]]+=dp[i-1][t];
                dp[i][flag[t][a[i]]]%=mod;
            }
        }
        ll ans=0;
        for(ll i=1;i<=maxx;i++)
        {
            ans+=(i*dp[x][i])%mod;
            ans%=mod;
        }
        printf("%I64d\n",ans);
    }
    return 0;
}