【URAL】1091 Tmutarakan Exams

题意：取k个不同的数，每个数不超过s，问种数。

若kx1,kx2,...,kx3满足条件，则x1,x2,...,x3必然满足条件。

因此枚举素数容斥，2*3*5*7>50，所以枚举之多三层。

#include<cstdio>
#include<cstring>
#include<vector>
#define MAXN 60
typedef long long LL;
using namespace std;
bool p[MAXN];
vector<int> prime;
LL C[MAXN][MAXN];
void Init() {
    int i, j;
    memset(p, true, sizeof(p));
    for (i = 2; i < 10; i++) {
        if (p[i]) {
            for (j = i * i; j < MAXN; j += i)
                p[j] = false;
        }
    }
    prime.clear();
    for (i = 2; i < MAXN; i++) {
        if (p[i])
            prime.push_back(i);
    }
    memset(C, 0, sizeof(C));
    for (i = 0; i < MAXN; i++)
        C[i][0] = C[i][i] = 1;
    for (i = 1; i < MAXN; i++) {
        for (j = 1; j < i; j++)
            C[i][j] = C[i - 1][j] + C[i - 1][j - 1];
    }
}
int main() {
    LL ans;
    int t, s, i, j, k;
    Init();
    while (~scanf("%d%d", &t, &s)) {
        for (ans = i = 0; i < (int) prime.size(); i++)
            ans += C[s / prime[i]][t];
        for (i = 0; i < (int) prime.size(); i++) {
            for (j = i + 1; j < (int) prime.size(); j++)
                ans -= C[s / prime[i] / prime[j]][t];
        }
        for (i = 0; i < (int) prime.size(); i++) {
            for (j = i + 1; j < (int) prime.size(); j++) {
                for (k = j + 1; k < (int) prime.size(); k++)
                    ans += C[s / prime[i] / prime[j]
                            / prime[k]][t];
            }
        }
        if (ans > 10000)
            ans = 10000;
        printf("%lld\n", ans);
    }
    return 0;
}