2017杭電多校聯賽第二場-TrickGCD （hdu6053）莫比烏斯容斥

TrickGCD

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0

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 kth 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

1 4 4 4 4 4

 

Sample Output

Case #1: 17

題目大意：求有多少組可能使得滿足題目要求

先來解釋樣例：比4不大的滿足條件的有2,3,4，而如果要區間內有共同的最大公約數的話就是2,4，樣例有4個

4，所以每一個有2種可能，即有2*2*2*2=16種，然後還有一種3,3,3,3，最大公約數爲3，所以有16+1=17

（種）所以我們可以想到的是容斥原理，2的倍數的加上3的倍數的減去2*3的倍數的，同理，加上5的倍數的減

去2*5的倍數的，減去3*5的倍數的，再加上2*3*5的倍數的，所以是莫比烏斯容斥，具體看代碼詳解：

ac代碼：

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<iostream>
#include<algorithm>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<map>
#include<string>
#define nl n<<1
#define nr (n<<1)|1
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int>P;
const int INF=0x3f3f3f3f;
const ll INFF=0x3f3f3f3f3f3f3f3f;
const double pi=acos(-1.0);
const double eps=1e-9;
const ll mod=1e9+7;
ll mu[100005];
void mobius()
{
    mu[1]=1;
    for(int i=1;i<=100000;i++)
        for(int j=i+i;j<=100000;j+=i)
            mu[j]-=mu[i];
}
ll qpow(ll a,ll b)
{
    ll ans=1;
    while(b)
    {
        if(b&1)
            ans=(ans*a)%mod;
        b>>=1;
        a=(a*a)%mod;
    }
    return ans;
}
ll sum[200005];
ll cnt[200005];
int main()
{
    mobius();int ca=1;
    int t;scanf("%d",&t);
    while(t--)
    {
        memset(cnt,0,sizeof(cnt));int n;
        scanf("%d",&n);int maxx=INF;
        for(int i=1;i<=n;i++)
        {
            int x;scanf("%d",&x);
            cnt[x]++;
            maxx=min(maxx,x);
        }
        ll ans=0;
        for(int i=1;i<=200000;i++)sum[i]=sum[i-1]+cnt[i];
        for(int i=2;i<=maxx;i++)
        {
            ll tmp=1;
            for(int j=1;j*i<=100000;j++)
            {
                tmp=(tmp*qpow((ll)j,sum[i*j+i-1]-sum[i*j-1]))%mod;
            }
            ans=(ans-tmp*mu[i]+mod)%mod;
        }
        printf("Case #%d: %lld\n",ca++,ans);
    }
    return 0;
}

題目鏈接：點擊打開鏈接http://acm.hdu.edu.cn/showproblem.php?pid=6053