Ultraviolet A 4683-find the number

Ultraviolet A 4683

This question refers to a set with a maximum of 12 elements. Find the nth number that can only be divided by one of the sets and only one number. (N <= 10 ^ 15 ).

I did this by using the refresh principle. First, we will accumulate the number of elements that can be divisible by each number. Of course, there will be duplicates. If a number is composed of two numbers in a set, it must be an integer multiple of all the numbers minus two times because it is the common number of the two numbers, when a certain number is three of them, it must also be two of them. In this way, C [k] [2] is reduced, and the number must be added. And so on.

The N number is required. The answer is 10 ^ 15 at the maximum. I divide the number by 10 ^ 15. If the number of matching conditions in [1, m] is res, if res> = N, high = mid-1; otherwise, low = Mid + 1.

But I have not passed the code, no limit to TLE .. The code will be posted here first, hoping that the passing experts will give you some advice.

#include <stdio.h>#include <iostream>#include <map>#include <set>#include <bitset>#include <list>#include <stack>#include <vector>#include <math.h>#include <string.h>#include <queue>#include <string>#include <stdlib.h>#include <algorithm>//#define long long __int64//#define LL long long#define eps 1e-9#define PI acos(-1.0)//typedef __int64 LL;using namespace std;const long long Max = 1000000000000000;int k;long long n;int a[15];long long C[15][15];long long gcd(long long a, long long b){    if(b == 0) return a;    return gcd(b,a%b);}long long lcm(long long a, long long b){    return a/gcd(a,b)*b;}void init(){    memset(C,0,sizeof(C));    for(int i = 1; i <= 12; i++)    {        for(int j = 0; j <= i; j++)        {            if(j == 0 || j == i)                C[i][j] = 1;            else if(j == 1)                C[i][j] = i;            else                C[i][j] = C[i-1][j-1] + C[i-1][j];        }    }}long long cal(long long m){    long long ans = 0;    for(int i = 1; i < (1<<k); i++)    {        int cnt = 0;        long long mul = 1;        for(int j = 0; j < k; j++)        {            if(i&(1<<j))            {                cnt++;                mul = lcm(mul,a[j]);            }        }        if(cnt&1)            ans += C[k][cnt-1]*(m/mul);        else            ans -= cnt*(m/mul);    }    return ans;}int main(){    int test;    init();    scanf("%d",&test);    while(test--)    {        scanf("%d %I64d",&k,&n);        for(int i = 0; i < k; i++)        {            scanf("%d",&a[i]);        }        long long low = 0,high = Max;        while(low <= high)        {            long long mid = (low + high)/2;            long long res = cal(mid);            if(res >= n)                high = mid-1;            else                low = mid+1;        }        printf("%I64d\n",low);    }    return 0;}

Ultraviolet A 4683-find the number