HDU 1796How many integers can you find(简单容斥定理)

HDU 1796How many integers can you find(简单容斥定理)

分类： 数论2013-10-12
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容斥

How many integers can you find

Time Limit: 12000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 3315 Accepted Submission(s): 937



Problem Description

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.

Input

There are a lot of cases. For each case, the first line contains two integers N and M. The follow line contains the M integers, and all of them are different from each other. 0<N<2^31,0<M<=10, and the M integer are non-negative and won’t exceed 20.

Output

For each case, output the number.

Sample Input

12 2
2 3

Sample Output

7

Author

wangye

Source

2008
“Insigma International Cup” Zhejiang Collegiate Programming Contest - Warm Up（4）

题目大意：很简单的题目，直接看意思就懂哈！

解题思路：容斥定理，加奇减偶，开始忘记求lcm了，囧！！而且开始还特判0的情况，题目中说的必须是除以，所以0不是一个解。。。开始竟然以为需要是因子就可以了。想通了之后直接先筛选一次，把0都筛选出去。

题目地址：How many integers can you find

AC代码：

[cpp] view
plaincopy

#include<iostream>

#include<cstring>

#include<string>

#include<cmath>

#include<cstdio>

using namespace std;

__int64 sum;

int n,m;

int a[25];

int b[25];

int visi[25];

__int64 gcd(__int64 m,__int64 n)

{

__int64 tmp;

while(n)

{

tmp=m%n;

m=n;

n=tmp;

}

return m;

}

__int64 lcm(__int64 m,__int64 n)

{

return m/gcd(m,n)*n;

}

void cal()

{

int flag=0,i;

__int64 t=1;

__int64 ans;

for(i=0;i<m;i++)

{

if(visi[i])

{

flag++; //记录用了多少个数

t=lcm(t,b[i]);

}

}

ans=n/t;

if(n%t==0) ans--;

if(flag&1) sum+=ans; //加奇减偶

else sum-=ans;

}

int main()

{

int i,j,p;

while(~scanf("%d%d",&n,&m))

{

sum=0;

for(i=0;i<m;i++)

scanf("%d",&a[i]);

int tt=0; //

for(i=0;i<m;i++)

{

if(a[i]) //去掉0

b[tt++]=a[i];

}

m=tt;

p=1<<m; //p表示选取多少个数，组合数的状态

for(i=1;i<p;i++)

{

int tmp=i;

for(j=0;j<m;j++)

{

visi[j]=tmp&1;

tmp>>=1;

}

cal();

}

printf("%I64d\n",sum);

}

return 0;

}

/*

12 2

2 3

12 3

2 3 0

12 4

2 3 2 0

*/

//968MS