hdu 5656 dp

CA Loves GCD

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

Total Submission(s): 1212    Accepted Submission(s): 397

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

222 431 2 3

 

Sample Output

810

 

Source

BestCoder Round #78 (div.2)

看了题解才知道是个dp 比赛中一直以为是数论题。。。（不过有大神用了容斥做的）

题目要求给你n个数问所有组合的最大公约数求和%10e8+7

令dp[i][j] 为前i个数取若干个数的gcd为j的方法数

dp[i][j]+=dp[i-1][j]  前i个数肯定包含前i-1个数的方法

当我们加入第i个数a[i] 会对原先的状态产生什么影响呢？

dp[i][gcd(a[i],j)]+=dp[i-1][j]  前i-1个数gcd为j的方法数有dp[i-1][j]个 a[i]与j 产生的gcd的方法数可以再加上dp[i-1][j] 个

#include <bits/stdc++.h>using namespace std;#define ll long longconst int Mod=100000007;int gcd(int a,int b){    return b==0?a:gcd(b,a%b);}int a[1005];int GCD[1005][1005];__int64 dp[1005][1005];void gg(){    memset(GCD,0,sizeof(GCD));    for(int i=1;i<1005;i++)        for(int k=1;k<1005;k++)            GCD[i][k]=GCD[k][i]?GCD[k][i]:gcd(k,i);}int main(){    int t;    gg();    scanf("%d",&t);    while(t--)    {        int n;        int tem=0;        scanf("%d",&n);        memset(dp,0,sizeof(dp));        for(int i=1;i<=n;i++)        {            scanf("%d",a+i);            tem=max(tem,a[i]);        }        for(int i=1;i<=n;i++)        {            for(int j=1;j<=tem;j++)            {                dp[i][j]=(dp[i][j]+dp[i-1][j]+(a[i]==j))%Mod; //当a[i] == j 时个数加1 记得取模                dp[i][GCD[a[i]][j]]=(dp[i][GCD[a[i]][j]]+dp[i-1][j])%Mod;            }        }        ll res=0;        for(ll j=1;j<=tem;j++)        {            res=(res+dp[n][j]*j)%Mod;        }        printf("%lld\n",res);    }    return 0;}