【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。

 1 #include<cstdio>

 2 #include<cstring>

 3 #include<vector>

 4 #define MAXN 60

 5 typedef long long LL;

 6 using namespace std;

 7 bool p[MAXN];

 8 vector<int> prime;

 9 LL C[MAXN][MAXN];

10 void Init() {

11     int i, j;

12     memset(p, true, sizeof(p));

13     for (i = 2; i < 10; i++) {

14         if (p[i]) {

15             for (j = i * i; j < MAXN; j += i)

16                 p[j] = false;

17         }

18     }

19     prime.clear();

20     for (i = 2; i < MAXN; i++) {

21         if (p[i])

22             prime.push_back(i);

23     }

24     memset(C, 0, sizeof(C));

25     for (i = 0; i < MAXN; i++)

26         C[i][0] = C[i][i] = 1;

27     for (i = 1; i < MAXN; i++) {

28         for (j = 1; j < i; j++)

29             C[i][j] = C[i - 1][j] + C[i - 1][j - 1];

30     }

31 }

32 int main() {

33     LL ans;

34     int t, s, i, j, k;

35     Init();

36     while (~scanf("%d%d", &t, &s)) {

37         for (ans = i = 0; i < (int) prime.size(); i++)

38             ans += C[s / prime[i]][t];

39         for (i = 0; i < (int) prime.size(); i++) {

40             for (j = i + 1; j < (int) prime.size(); j++)

41                 ans -= C[s / prime[i] / prime[j]][t];

42         }

43         for (i = 0; i < (int) prime.size(); i++) {

44             for (j = i + 1; j < (int) prime.size(); j++) {

45                 for (k = j + 1; k < (int) prime.size(); k++)

46                     ans += C[s / prime[i] / prime[j]

47                             / prime[k]][t];

48             }

49         }

50         if (ans > 10000)

51             ans = 10000;

52         printf("%lld\n", ans);

53     }

54     return 0;

55 }