HDU 6053 TrickGCD（枚举）-优快云博客

TrickGCD

Time Limit: 5000/2500 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 2855 Accepted Submission(s): 1075

Problem Description

You are given an array

A , and Zhu wants to know there are how many different array

B satisfy the following conditions?

*

1≤Bi≤Ai
* For each pair( l , r ) (

1≤l≤r≤n ) ,

gcd(bl,bl+1...br)≥2

Input

The first line is an integer T(

1≤T≤10 ) describe the number of test cases.

Each test case begins with an integer number n describe the size of array

A .

Then a line contains

n numbers describe each element of

A

You can assume that

1≤n,Ai≤105

Output

For the

k th test case , first output "Case #k: " , then output an integer as answer in a single line . because the answer may be large , so you are only need to output answer

mod

109+7

Sample Input

Sample Output

  
   Case #1: 17

Source

2017 Multi-University Training Contest - Team 2

题意：

给你a数列，求出gcd>=2的b数列数量，b[i]每个数都小于等于a[i]。

POINT：

枚举gcd，然后容斥。记录a[i]数组的数量，然后枚举。

比如枚举2，就看a[i]里有多少2和3。具体看代码。

#include <stdio.h>
#include <iostream>
#include <string.h>
#include <math.h>
#include <algorithm>
#include <map>
using namespace std;
#define LL long long
const int N = 1e5;
const LL p= 1e9+7;
LL pre[N+4],sum[N+4];
LL num[N+4];
LL qkm(LL base,LL mi)
{
    base%=p;
    LL ans=1;
    while(mi)
    {
        if(mi&1) ans=ans*base;
        base*=base;
        mi>>=1;
        base%=p;
        ans%=p;
    }
    return ans;
}
int main()
{
    int T;
    int k=0;
    scanf("%d",&T);
    while(T--)
    {
        memset(pre,0,sizeof pre);
        memset(num,0,sizeof num);
        memset(sum,0,sizeof sum);
        int n,o;
        scanf("%d",&n);
        int maxx=0;
        for(int i=1;i<=n;i++)
            scanf("%d",&o),pre[o]++,maxx=max(maxx,o);
        for(int i=1;i<=N;i++) sum[i]=sum[i-1]+pre[i];
        int flag=1;
        for(int i=2;i<=N;i++)
        {
            if(!flag) break;
            else
            {
                num[i]=1;
                for(int j=0;j<=maxx;j+=i)
                {
                    int a=j/i;
                    LL b=sum[min(j+i-1,N)]-sum[max(j-1,0)];
                    if(!a&&b)
                    {
                        num[i]=0;
                        flag=0;
                        break;
                    }
                    num[i]=(num[i]*qkm(a,b))%p;
                }
            }
        }
        for(int i=N;i>=2;i--)
        {
            for(int j=2*i;j<=N;j+=i)
            {
                num[i]-=num[j];
                num[i]=(num[i]+p)%p;
            }
        }
        LL ans=0;
        for(int i=2;i<=N;i++)
        {
            (ans+=num[i])%=p;
        }
        printf("Case #%d: %lld\n",++k,ans);
        
    }

    
}