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\n",x)
 #define pd(x) printf("%f\n",x)
 #define pl(x) printf("%lld\n",x)
 #define fr(i,a,b) for(int i=a;i<=b;i++)
 using namespace std;
 const int N=1e5+;
 const int mod=1e9+;
 const int MOD=mod-;
 const db  eps=1e-;
 const int inf = 0x3f3f3f3f;
 ll a[N];
 ll sum[N];
 ll qpow(ll x,ll n)
 {
     ll ans=;x%=mod;
     for(;n>;n>>=){if(n&) 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=;ii<=t;ii++)
     {
         int n,x,ma=-N,mi=N;
         memset(sum,,sizeof(sum));
         ci(n);
         for(int i=;i<n;i++)
             ci(x),sum[x]++,ma=max(ma,x),mi=min(mi,x);
         for(int i=;i<=ma;i++) sum[i]+=sum[i-];//统计小于等于i的数字的个数
         for(int i=;i<=mi;i++){//从2～mi枚举gcd
             a[i]=;
             for(int j=i;j<=ma;j+=i){//分区间计算
                 int c;
                 if(i+j-<=ma) c=sum[i+j-]-sum[j-];
                 else c=sum[ma]-sum[j-];
                 if(!c) continue;
                 a[i]=(a[i]*qpow(j/i,c))%mod;
             }
         }
         ll ans=;
         for(int i=mi;i>=;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\n",ii,ans);
     }
 }