HDU-6053 TrickGCD

题目连接：

https://vjudge.net/problem/HDU-6053

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≤10^5

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

Hint

题意

给出序列A，求出满足条件的序列B，使得：

$ 1 <= Bi <= Ai \(
\)gcd(B_1, B_2,B_3\cdots,B_n) >= 2$

题解：

考虑用总的序列B个数减去gcd为1的序列B的个数，就是gcd大于等于2的个数。

但是原本计算G[i]的式子由O(1)变成了O(n),也没法进行分块了，复杂度变成了O(n^2)

可以把原本计算G[i]的式子\(mu[i] * \prod_{j=1}^n\frac{B_j}{i}\)

简化为\(mu[i] * \prod_{j=1}^{\frac{maxA[1-n]}{i}} j^{count(j)}\)

这里的count(j)是A序列中除以i等于j的数的个数

设sum[i]为A序列中小于等于i的数的个数则count(j) = sum[i(j+1)-1]-sum[ij-1]

于是整体复杂度变为\(\sum_1^n\frac{maxA[1-n]}{i} = nlogn\)

代码

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int mx = 1e5+5;
const int mod = 1e9+7;
const int INF = 0x3f3f3f3f;
typedef long long ll;

bool vis[mx];
int sum[mx], prim[mx], mu[mx], num[mx], cnt = 0;
int a[mx];

void get_mu(){
    mu[1] = 1;
    for(int i = 2; i< mx; i++){
        if(!vis[i]) {
            prim[++cnt] = i;
            mu[i] = -1;
            num[i] = 1;
        }
        for(int j = 1; j <= cnt && prim[j]*i < mx; j++) {
            vis[prim[j]*i] = 1;
            num[prim[j]*i] = num[i] + 1;
            if(i % prim[j] == 0) break;
            else mu[i*prim[j]] = -mu[i];
        }
    }
}

ll pow_mod(ll a, ll b) {
    ll ans = 1;
    while (b > 0) {
        if (b & 1) ans = ans * a % mod;
        a = a * a % mod;
        b /= 2;
    }
    return ans;
}

int main() {
    // freopen("in.txt", "r", stdin);
    // freopen("out1.txt", "w+", stdout);
    get_mu();
    int T, kase = 0;
    scanf("%d", &T);

    while (T--) {
        memset(a, 0, sizeof(a));
        memset(sum, 0, sizeof(sum));
        int n;
        scanf("%d", &n);
        ll tot = 1;
        int len = INF, mlen = 0;
        for (int i = 1; i <= n; i++) {
            int x;
            scanf("%d", &x);
            a[x]++;
            len = min(len, x);
            mlen = max(mlen, x);
            tot = tot * x % mod;
        }
        for (int i = 1; i < mx; i++) sum[i] = sum[i-1] + a[i];
        ll ans = 0;

        for (int i = 1; i <= len; i++) {
            ll tmp = mu[i];
            for (int j = 1; j <= mlen/i; j++) {//计算公约数为i的个数
                tmp *= pow_mod(j, sum[min(i*(j+1)-1, mlen)]-sum[i*j-1]);
                tmp %= mod;
            }
            ans += tmp;
            ans = (ans % mod + mod) % mod;
        }
        ans = ((tot-ans) % mod + mod) % mod;
        printf("Case #%d: %lld\n", ++kase, ans);
    }

    return 0;
}