HDU 6053 TrickGCD —— 2017 Multi-University Training 2

TrickGCD

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 2523    Accepted Submission(s): 965

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≤1e5

 

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 1e9+7

 

Sample Input

1
4

4 4 4 4

 

Sample Output

Case #1: 17

 

题目大意：有数组A，根据1<=Bi<=Ai,且对于任意1<=l<=r<=n, gcd(Bl,Bl+1...Br)≥2 构造B数列

思路：B的最小值一定不会大于A的最小值，所以gcd(B1,B2,...,Bn)一定不大于A的最小值，那么我们可以求出A的最小值，对gcd为质数的情况求和，利用容斥原理减去重复的情况。在计算重复情况时，发现对于gcd是奇数个质数的乘积时是加，偶数个时是减，与莫比乌斯函数相反，那么就可以借助莫比乌斯函数求和。这里暴力求和会超时，问了学长之后才知道分桶计算可以快很多，涨姿势了。

 

AC代码：

 #include<iostream>
 #include<cstring>
 #include<cstdio>
 using namespace std;
 typedef long long LL;
 const int MAXN=1e5+;
 const int MOD=1e9+;
 LL miu[MAXN],primes[MAXN],tot=;
 bool isprime[MAXN];
 int a[MAXN],sum[MAXN];
 int maxx=,minn=MAXN;
 void getmiu()
 {
     memset(isprime, , sizeof((isprime)));
     miu[]=;
     for(int i=;i<MAXN;i++){
         if(isprime[i]==false)
         {
             miu[i]=-;
             primes[++tot]=i;//cout<<'*'<<endl;
         }
         for(int j=;j<=tot;j++){
             if(i*primes[j]>=MAXN) break;
             isprime[i*primes[j]]=;
             if(i%primes[j]==){
                 miu[i*primes[j]]=;
                 break;
             }
             miu[i*primes[j]]=-miu[i];
         }
     }

     for(int i=;i<MAXN;i++) miu[i]=-miu[i];
 }
 LL quick_pow(LL a, LL p)
 {
     int res=;
     while(p)
     {
         if(p&) res=a*res%MOD;
         a=a*a%MOD;
         p>>=;
     }
     return res;
 }
 int solve()
 {
     LL i,j,k,p;
     LL ans=;
     for(int i=;i<=minn;i++){
         if(!miu[i]) continue;
         LL res=;
         j=min(i, maxx), k=min((i<<)-, maxx);
         for(p=; ;p++)
         {
             if(sum[k]-sum[j-])
                 res=res*quick_pow(p, sum[k]-sum[j-])%MOD;
             if(k==maxx) break;
             j+=i;
             k+=i;
             if(k>maxx) k=maxx;
         }
         ans+=miu[i]*res;
         if(ans>MOD) ans-=MOD;
         if(ans<) ans+=MOD;
     }
     return ans%MOD;

 }
 int main()
 {
     int T,n,t=;
     getmiu();
     scanf("%d", &T);
     //cout<<'*'<<endl;
     while(T--)
     {
         scanf("%d", &n);
         maxx=,minn=MAXN;
         memset(sum, , sizeof(sum));
         for(int i=;i<n;i++){
             scanf("%d", &a[i]);
             sum[a[i]]++;
             maxx=max(a[i], maxx);
             minn=min(a[i], minn);
         }
         for(int i=;i<=maxx;i++) sum[i]+=sum[i-];
         LL res=solve();
         printf("Case #%d: %lld\n", ++t, res);
     }
 }