HDU6053-TrickGCD

TrickGCD

                                                                      Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
                                                                                                Total Submission(s): 697    Accepted Submission(s): 266

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

 

Source

2017 Multi-University Training Contest - Team 2

 

题意：给你n个数字，每个位置的数字可以小于等于a[i]，求所有gcd(l,r)都满足大于等于2的情况数

解题思路：枚举gcd的情况，每种gcd的情况等于所有a[i]/gcd的乘积，这个需要优化，枚举除数，a[i]/gcd相同的为一块（现场的时候没想到这个），然后重复的用容斥搞一下就好了

#include <iostream>  #include <cstdio>  #include <cstring>  #include <string>  #include <algorithm>  #include <map>  #include <cmath>#include <set>  #include <stack>  #include <queue>  #include <vector>  #include <bitset>  #include <functional>using namespace std;#define LL long long  const int INF = 0x3f3f3f3f;const LL mod = 1000000007;int sum[100009], n;LL x[100009];LL qpow(LL x, LL y){    LL ans = 1;    while (y)    {        if (y & 1) ans = (ans*x) % mod;        y >>= 1;        x = (x*x) % mod;    }    return ans;}int main(){    int t, cas = 1;    scanf("%d", &t);    while (t--)    {        memset(sum, 0, sizeof sum);        scanf("%d", &n);        for (int i = 1; i <= n; i++)        {            int a;            scanf("%d", &a);            sum[a]++;        }        for (int i = 1; i <= 100005; i++) sum[i] += sum[i - 1];        for (int i = 2; i <= 100000; i++)        {            x[i] = 1LL;            for (int j = 0; j <= 100000; j += i)            {                int a, cnt;                if (j + i - 1 > 100000) cnt = sum[100000] - sum[j - 1];                else if (j == 0) cnt = sum[j + i - 1];                else cnt = sum[j + i - 1] - sum[j - 1];                a = j / i;                if (a == 0 && cnt) x[i] = 0;                else if (cnt) x[i] = (x[i] * qpow(a, cnt)) % mod;            }        }        for (int i = 100000; i >= 2; i--)        {            for (int j = i + i; j <= 100000; j += i)                x[i] -= x[j], x[i] = (x[i] % mod + mod) % mod;        }        LL ans = 0;        for (int i = 2; i <= 100000; i++) ans += x[i], ans %= mod;        printf("Case #%d: %lld\n", cas++, ans);    }    return 0;}