HDU 5656 CA Loves GCD

CA Loves GCD

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

Total Submission(s): 1730    Accepted Submission(s): 550
[/b]

[align=left]Problem Description[/align]
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.

 

[align=left]Input[/align]
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
 

[align=left]Output[/align]
T
lines, each line prints the sum of GCDs mod 100000007.
 

[align=left]Sample Input[/align]

2
2
2 4
3
1 2 3

 

[align=left]Sample Output[/align]

8
10

 

[align=left]Source[/align]
BestCoder Round #78 (div.2)

 

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题意：n个数里面任意挑选几个数出来，求剩下的数的gcd之和。因为范围在1000之间，我们可以计算出gcd为1到1000的所有方案数。

令dp[i][j]代表选了前i个数 gcd为j的方案数。选i+1个数转移为dp[i][j]+=dp[i+1][f[j][a[i+1]]] 。不选为dp[i][j]+=dp[i+1][j];其中f[i][j]代表i，j两数的gcd

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <string>
#include <cstdio>
#include <cmath>
using namespace std;
typedef long long LL;
int f[1001][1001];
int dp[1001][1001];
int a[1001];
const int mod=100000007;
inline void add(int b,int &a)
{
if((a+=b)>=mod)a-=mod;
}
int main()
{
int t;
int i,j;
for(i=0; i<=1000; i++)
for(j=0; j<=1000; j++)
f[i][j]=__gcd(i,j);
scanf("%d",&t);
while(t--)
{
int n;
cin>>n;
memset(a,0,sizeof(a));
for(i=1; i<=n; i++)
cin>>a[i];
memset(dp,0,sizeof(dp));
dp[0][0]=1;
for(i=0; i<n; i++)
for(j=0; j<=1000; j++)
{
if(dp[i][j])
add(dp[i][j],dp[i+1][j]);
add(dp[i][j],dp[i+1][f[j][a[i+1]]]);
}
LL ans=0;
for(i=1; i<=1000; i++)
ans+=LL (i)*dp
[i];
printf("%lld\n",ans%mod);
}
return 0;
}