HDU 6053 ( TrickGCD ) 分块＋容斥

TrickGCD

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 3401    Accepted Submission(s): 1268

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

 

Source

2017 Multi-University Training Contest - Team 2

 

思路：枚举gcd ，对于当前的i，分区间地计算出gcd为i倍数的数列总数，之后利用容斥来减掉重复的部分，要保证每次减掉都是确定的倍数因此要倒着减。

代码：

#include<bits/stdc++.h>
#define db double
#define ll long long
#define ci(x) scanf("%d",&x)
#define cd(x) scanf("%lf",&x)
#define cl(x) scanf("%lld",&x)
#define pi(x) printf("%d
",x)
#define pd(x) printf("%f
",x)
#define pl(x) printf("%lld
",x)
#define fr(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
const int N=1e5+5;
const int mod=1e9+7;
const int MOD=mod-1;
const db  eps=1e-10;
const int inf = 0x3f3f3f3f;
ll a[N];
ll sum[N];
ll qpow(ll x,ll n)
{
    ll ans=1;x%=mod;
    for(;n>0;n>>=1){if(n&1) ans=(ans*x)%mod;x=x*x%mod;}
    return ans;
}
int main()
{
//    ios::sync_with_stdio(false);
//    cin.tie(0);
    int t;
    ci(t);
    for(int ii=1;ii<=t;ii++)
    {
        int n,x,ma=-N,mi=N;
        memset(sum,0,sizeof(sum));
        ci(n);
        for(int i=0;i<n;i++)
            ci(x),sum[x]++,ma=max(ma,x),mi=min(mi,x);
        for(int i=1;i<=ma;i++) sum[i]+=sum[i-1];//统计小于等于i的数字的个数
        for(int i=2;i<=mi;i++){//从2～mi枚举gcd
            a[i]=1;
            for(int j=i;j<=ma;j+=i){//分区间计算
                int c;
                if(i+j-1<=ma) c=sum[i+j-1]-sum[j-1];
                else c=sum[ma]-sum[j-1];
                if(!c) continue;
                a[i]=(a[i]*qpow(j/i,c))%mod;
            }
        }
        ll ans=0;
        for(int i=mi;i>=2;i--)
        {
            for(int j=i+i;j<=mi;j+=i) a[i]=(a[i]-a[j])%mod;//减掉确定的倍数
            a[i]=(a[i]+mod)%mod;
            ans=(ans+a[i])%mod;
        }
        printf("Case #%d: %lld
",ii,ans);
    }
}