HDU - 1796 How many integers can you find （容斥）

题目链接点击打开链接

 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 (非负数)
and won’t exceed

这个题需要注意的有三点：

1，注意0的处理（把0 略去）     (因为这个RE了一直）

2，注意应该是最小公倍数           
(WA!!!!)

3，long long

先写上错误代码；

然后写出错误原因：很简单的一个栗子：

13 2
4 6

正确答案是4；
而这个程序的答案是5；

错误在于它把12 算了两次还没减去；

#include <iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<vector>

using namespace std;

long long n;
long long s[100];
vector<long long > q;
long long solve(int num)
{
long long ans=0;
for(long long i=1;i<(1<<num);i++)
{
long long p=1;
int cnt=0;
for(long long j=0;j<num;j++)
{
// cout<<((1<<j)&i)<<endl;
if((1<<j)&i)
{
p*=q[j];
cnt++;
}
}
//cout<<cnt<<endl;
if(cnt&1)
ans+=(n-1)/p;
else
ans-=(n-1)/p;
}
return a
4000
ns;
}
int main()
{
int num=0;
while(cin>>n>>num)
{
q.clear();
int flag=0;
for(int i=0;i<num;i++)
{
cin>>s[i];
}
sort(s,s+num);
for(int i=0;i<num;i++)
{
if(s[i]<=0)
continue;
for(int j=i+1;j<num;j++)
{
if(s[j]%s[i]==0)
s[j]=-1;
}
q.push_back(s[i]);
}
// cout<<q.size()<<endl;
long long ans=solve(q.size());
cout<<ans<<endl;
}
return 0;
}


所以正确的打开方式应该是这样子的：

#include <iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<vector>

using namespace std;

long long n;
long long s[100];
vector<long long > q;
int gcd(int a,int b)
{
if(b==0)
return a;
else
return gcd(b,a%b);
}
long long  solve(int num)
{
long long ans=0;
for(long long i=1;i<(1<<num);i++)
{
long long p=1;
int cnt=0;
for(long long  j=0;j<num;j++)
{
// cout<<((1<<j)&i)<<endl;
if((1<<j)&i)
{
p=p/gcd(p,q[j])*q[j];//求当前组合数的最小公倍数
cnt++;
}
}
//cout<<cnt<<endl;
if(cnt&1)
ans+=(n-1)/p;
else
ans-=(n-1)/p;
}
return ans;
}
int main()
{
int num=0;
while(cin>>n>>num)
{
q.clear();
int flag=0;
for(int i=0;i<num;i++)
{
cin>>s[i];
}
sort(s,s+num);
for(int i=0;i<num;i++)
{
if(s[i]<=0)
continue;
for(int j=i+1;j<num;j++)
{
if(s[j]%s[i]==0)
s[j]=-1;
}
q.push_back(s[i]);
}
//  cout<<q.size()<<endl;
long long ans=solve(q.size());
cout<<ans<<endl;
}
return 0;
}

呦呦切克闹容斥 我来了