HDU 6503 TrickGCD

TrickGCD

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

144 4 4 4

 

Sample Output

Case #1: 17

题意：给出数列 A[N]，问满足：
  1 <= B[i] <= A[i] ;
对任意(l, r) (1<=l<=r<=n) ，使得 gcd(bl,...br) >= 2 ;的 B[N] 数列的个数
分析：
设 gcd(b1,...bn) = k (k >= 2)，此时 k 对答案的贡献为 (a1/k)*(a2/k)*(a3/k)*...*(an/k)
根据容斥原理，ans = +[k=一个素数之积 时对答案的贡献]-[k=两个素数之积 时对答案的贡献]+[k=三个素数之积 时对答案的贡献]...
故任意k对答案的贡献系数 μ(k) = 0 , k是完全平方数的倍数 = (-1)^(n-1) , k = p1*p2*p3*...*pn ,p是素数
贡献系数可以O(nsqrt(n)) 或者 O(nlogn) 预处理，再或者可以看出μ(k) 是莫比乌斯函数的相反数
现在枚举k需要O(n)的时间，计算k对答案的贡献必须在O(sqrt(n))的时间之内
将a[]处理成权值数组，并求前缀和，设为 sum[]
对于每个k，对sum[]进行埃式筛法的分块，即根据k的倍数分块
此时每个k的贡献 = 1^(sum[2k-1]-sum[k-1]) * 2^(sum[3k-1]-sum[2k-1]) * 3^(sum[4k-1]-sum[3k-1]) ...
就做到 O(n(logn)^2)

代码如下：

#include  <bits/stdc++.h>using namespace std;#define LL long longconst LL MOD = 1e9+7;const int N = 1e5+4;bool notp[N];int prime[N], pnum, mu[N];void Mobius() {//用莫比乌斯素数打表 memset(notp, 0, sizeof(notp));mu[1] = 1;for (int i = 2; i < N; i++) {if (!notp[i]) prime[++pnum] = i, mu[i] = -1;for (int j = 1; prime[j]*i < N; j++) {notp[prime[j]*i] = 1;if (i%prime[j] == 0) {mu[prime[j]*i] = 0;break;}mu[prime[j]*i] = -mu[i];}}for (int i = 0; i < N; i++) mu[i] = -mu[i];}LL PowMod(LL a, int m)//快速幂{    if (a == 1 || m == 0) return 1;    if (a == 0) return 0;    LL res = 1;    while (m)    {        if (m&1) res = res*a % MOD;        a = a*a % MOD;        m >>= 1;    }    return res;}int a[N];int t, n;int sum[N];int Max, Min;LL ans;void solve(){    int i, j, k, p;    ans = 0;    for (i = 2; i <= Min; ++i)    {        if (!mu[i]) continue;        LL res = 1;        j = min(i, Max), k = min((i<<1)-1, Max);        for (p = 1; ; ++p)        {            if (sum[k] - sum[j-1])                res = res*PowMod(p, sum[k] - sum[j-1]) % MOD;            if (k == Max) break;            j += i;            k += i;            if (k > Max) k = Max;        }        ans += mu[i]*res;        if (ans > MOD) ans -= MOD;        if (ans < 0) ans += MOD;    }}int main(){    int i;    Mobius();    scanf("%d", &t);    for (int tt = 1; tt <= t; ++tt)    {        scanf("%d", &n);        for (i = 0; i < N; ++i) sum[i] = 0;        for (i = 1; i <= n; ++i)        {            scanf("%d", &a[i]);            ++sum[a[i]];        }        Max = Min = a[1];        for (i = 2; i <= n; ++i)        {            Max = max(Max, a[i]);            Min = min(Min, a[i]);        }        for (i = 1; i <= Max; ++i) sum[i] += sum[i-1];        solve();        printf("Case #%d: %lld\n", tt, ans);    }}