http://acm.timus.ru/problem.aspx?space=1&num=1091

从1~s中选出k个数，使得k个数的最大公约数大于1，问这样的取法有多少种。（2<=k <= s<=50）

同素数四元组问题类似，可以参考http://blog.csdn.net/u013081425/article/details/40653895

只不过这里是选出k个，不是4个。

#include <stdio.h>
#include <iostream>
#include <map>
#include <set>
#include <bitset>
#include <list>
#include <stack>
#include <vector>
#include <math.h>
#include <string.h>
#include <queue>
#include <string>
#include <stdlib.h>
#include <algorithm>
#define LL __int64
//#define LL long long
#define eps 1e-9
#define PI acos(-1.0)
using namespace std;
const int maxn = 100010;

int k,s;
int num[55];
int prime[] = {15,2,3,5,7,11,13,17,19,23,29,31,37,41,43,47};
int cnt[55];
LL c[55][55];

void init()
{
    memset(c,0,sizeof(c));
    for(int i = 1; i <= 25; i++)
    {
       for(int j = 0; j <= i; j++)
        {
            if(j == 0 || j == i)
                c[i][j] = 1;
            else if(j == 1)
                c[i][j] = i;
            else
                c[i][j] = c[i-1][j-1] + c[i-1][j];
        }
    }
}

void dfs(int st, int val, int num)
{
    if(val > 50) return;
    cnt[val] = num&1 ? 1 : -1;
    for(int i = st; i <= prime[0]; i++)
    {
        if(prime[i]*val > 50)
            break;
        dfs(i+1,val*prime[i],num+1);
    }
}

int main()
{
    init();
    memset(cnt,0,sizeof(cnt));
    dfs(1,1,0);
    while(~scanf("%d %d",&k,&s))
    {
        for(int d = 2; d <= s; d++)
        {
            num[d] =  s/d;
        }

        LL ans = 0;

        for(int d = 2; d <= s; d++)
        {
            if(num[d] < k)
                break;
            if(cnt[d])
            {
                ans += cnt[d]*c[num[d]][k];
            }
        }
        if(ans >= 10000)
            printf("10000\n");
        else
            printf("%I64d\n",ans);
    }
    return 0;
}

ural 1091. Tmutarakan Exams(容斥)