BestCoder Round #78 (div.2)1002 CA Loves GCD

CA Loves GCD

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

Total Submission(s): 617 Accepted Submission(s): 217



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)

DP问题从来都是一项玄学，很多情况下用DP一下就能解决的问题，但往往我们就是想不到，此题就是一个DP问题，当然在比赛的讨论中有人说这题暴力也能过……我相信这题暴力应该是过不了终端的。

首先我们先看看官方的题解：

1002

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][gcd(j,v)] += dp[i]j

复杂度O(N*MaxV) MaxV 为出现过的数的最大值

其实有O(MaxV *log(MaxV))的做法，我们考虑记f[i]表示从这些数中选择若干个数，使得他们的gcd是i的倍数的方案数。假如有K个数是i的倍数，则f[i]=2^K-1，再用g[i]表示从这些数中选择若干个数，使得他们的gcd是i的方案数，则g[i]=f[i] - g[j] (对于所有j是i的倍数)。

由调和级数可以得到复杂度为O(MaxV *log(MaxV))

题解上写的很清楚，小编在这里解释一下状态转移方程的含义：dp[i+1][j] 如果不取a[i+1]这个数，那么dp[i][j]就可以达到dp[i+1][j]的状态，如果取到了a[i+1]这个数，那么dp[i][j]就可以达到dp[i+1][gcd(j,a[i+1])]的状态。这就是状态的转移方程，为了防止超时，可以再对1000*1000这个范围内对gcd进行预处理一下。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 1000+5;
const int MOD = 100000007;
int a[maxn],dp[maxn][maxn];
int GCD[maxn][maxn];
int gcd(int x,int y)
{
if(x == 0) return y;
return gcd(y%x,x);
}
int main()
{
for(int i=0; i<=1000; i++)
for(int j=0; j<=1000; j++)
GCD[i][j] = gcd(i,j);
int T,n;
scanf("%d",&T);
while(T--)
{
memset(dp,0,sizeof(dp));
scanf("%d",&n);
int Max = 0;
for(int i=1; i<=n; i++)
{
scanf("%d",&a[i]);
Max = max(a[i],Max);
}
dp[0][0] = 1;
for(int i=0; i<n; i++)
{
for(int j=0; j<=Max; 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;
}
}
}
int ans = 0;
for(int i=0; i<=Max; i++)
ans = (ans + ((long long)dp
[i])*i%MOD) % MOD;
printf("%d\n",ans);
}
return 0;
}