从1到s选出k个数 他们的最大公约数大于1 求方案数
容斥 S(1)-S(2)+S(3) S(x)为选出k个数的公因子个数为x的数量
#include <cstdio>
#include <cmath>
#include <cstring>
using namespace std;
typedef long long LL;
const int maxn = 55;
int prime[maxn], vis[maxn];
int n, m;
int get_primes(int n)
{
int m = sqrt(n+0.5), c = 0;
for(int i = 2; i <= n; i++)
if(!vis[i])
{
prime[c++] = i;
for(int j = i*2; j <= n; j += i)
{
vis[j] = 1;
}
}
return c;
}
LL cm(int n, int m)
{
LL ans = 1;
for(int i = 1; i <= m; i++)
{
ans *= (LL)n--;
ans /= (LL)i;
}
return ans;
}
void dfs(int i, int x, int k, int m, LL& ans, int c)
{
if(i == c)
{
if(!k)
return;
if(k&1)
ans += cm(m/x, n);
else
ans -= cm(m/x, n);
return;
}
dfs(i+1, x*prime[i], k+1, m, ans, c);
dfs(i+1, x, k, m, ans, c);
}
LL cal(int c)
{
LL ans = 0;
dfs(0, 1, 0, m, ans, c);
if(ans > 10000)
ans = 10000;
return ans;
}
int main()
{
int c = get_primes(50);
//printf("%d\n", c);
while(scanf("%d %d", &n, &m) != EOF)
{
printf("%lld\n", cal(c));
}
return 0;
}URAL 1091. Tmutarakan Exams 容斥
原文:http://blog.csdn.net/u011686226/article/details/40889613