HDU

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 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;}

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;}

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