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 程式師世界 >> 編程語言 >> C語言 >> C++ >> C++入門知識 >> HDU 1796 How many integers can you find 容斥原理

HDU 1796 How many integers can you find 容斥原理

編輯:C++入門知識

HDU 1796 How many integers can you find 容斥原理


 

How many integers can you find

Time Limit: 12000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4848 Accepted Submission(s): 1388



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 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)
已知n和m,有m個元素,求小於n且是m個元素中任意元素的倍數的個數。 基礎容斥:ans=整除1個元素個數-整除2個元素個數+整除3個元素個數-整除4個元素個數+....
//904MS	1596K
#include
#include
int s[27],vis[27],sum,n,m,k;

int gcd(int a,int b)//最大公約數
{
    return b?gcd(b,a%b):a;
}

int lcm(int a,int b)//最小公倍數
{
    return a/gcd(a,b)*b;
}

void dfs(int x,int ans,int now)//x代表當前第幾個數,ans代表一共ans個數求lcm,now代表當前有now個數
{
    if(now==ans)
    {
        int a=1;
        for(int i=1;i<=k;i++)
            if(vis[i])a=lcm(a,s[i]);
        if(ans&1)sum+=(n-1)/a;
        else sum-=(n-1)/a;
        return ;
    }
    for(;x<=k;x++)
        if(!vis[x])
        {
            vis[x]=1;
            dfs(x+1,ans,now+1);
            vis[x]=0;
        }
}
int main()
{
    while(scanf(%d%d,&n,&m)!=EOF)
    {
        int a;
        k=0,sum=0;
        for(int i=1;i<=m;i++)
        {
            scanf(%d,&a);
            if(a)s[++k]=a;
        }
        for(int i=1;i<=k;i++)
        {
            memset(vis,0,sizeof(vis));
            dfs(1,i,0);
        }
        printf(%d
,sum);
    }
    return 0;
}


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