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B - How many integers can you find

时间:2019-08-15 22:42:22      阅读:52      评论:0      收藏:0      [点我收藏+]

标签:因此   mes   scanf   can   example   整数   +=   二进制   ring   

  Now you get a number N, and a M-integers set, you should find out how many integers which are small than N, that they can divided exactly by any integers in the set. For example, N=12, and M-integer set is {2,3}, so there is another set {2,3,4,6,8,9,10}, all the integers of the set can be divided exactly by 2 or 3. As a result, you just output the number 7.

题目大意:给你一个数n,和m个0~20的整数i,让你求1~n-1中是i的倍数的数有几个。

显然,这是一个容斥定理的题,求对1~(n-1)中所有m个i的倍数的数的个数。并且m<=10,因此我们用二进制枚举法就能够列出所有的情况。

需要注意的就是m个数,把这m个数先化成互质的。最后在套用模板即可。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
ll a[500];
int main()
{
    ll n, m;
    while(~scanf("%lld%lld",&n,&m))
    {
        
        int top = 0;
        for(int i = 0; i < m; ++ i)
        {
            ll x;
            cin >> x;
            if(x != 0)                  
                a[top++] = x;
        }
        ll ans = 0,sum,num;
        for(int i = 1; i < (1 << top); ++ i)
        {
            num=0,sum=1;
            for(int j = 0; j < top; ++ j)
            {
                if((i >> j) & 1)
                {
                    num++;
                    sum = sum * a[j] /(__gcd(sum, a[j]));
                }
            }
            if(num % 2)
            {
                ans += (n - 1) / sum;   
            }
            else
            {
                ans -= (n - 1) / sum;
            }
        }
        cout << ans << endl;
    }
    return 0;
}

以上。

B - How many integers can you find

标签:因此   mes   scanf   can   example   整数   +=   二进制   ring   

原文地址:https://www.cnblogs.com/zjydeoneday/p/11360882.html

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