1091. Tmutarakan Exams
Time limit: 1.0 second
Memory limit: 64 MB
Memory limit: 64 MB
University
of New Tmutarakan trains the first-class specialists in mental
arithmetic. To enter the University you should master arithmetic
perfectly. One of the entrance exams at the Divisibility Department is
the following. Examinees are asked to find K different numbers that have a common divisor greater than 1. All numbers in each set should not exceed a given number S. The numbers K and S
are announced at the beginning of the exam. To exclude copying (the
Department is the most prestigious in the town!) each set of numbers is
credited only once (to the person who submitted it first).
Last year these numbers were K=25 and S=49
and, unfortunately, nobody passed the exam. Moreover, it was proved
later by the best minds of the Department that there do not exist sets
of numbers with the required properties. To avoid embarrassment this
year, the dean asked for your help. You should find the number of sets
of K different numbers, each of the numbers not exceeding S,
which have a common divisor greater than 1. Of course, the number of
such sets equals the maximal possible number of new students of the
Department.
Input
The input contains numbers K and S (2 ≤ K ≤ S ≤ 50).
Output
You
should output the maximal possible number of the Department's new
students if this number does not exceed 10000 which is the maximal
capacity of the Department, otherwise you should output 10000.
Sample
| input | output |
|---|---|
3 10 |
11 |
Problem Author: Stanislav Vasilyev
Problem Source: USU Open Collegiate Programming Contest March'2001 Senior Session
题目大意：求k个数的公约数 > 1有多少对，这些数不会超过S；
思路：由于数很少我们可以直接判断出有哪些数的倍数是两个约数的公倍数
2， 3共同的约数为6
2， 5 ~10
2， 7 ~14
2， 11 ~22
3 ， 5~15
3， 7 ~21
然后再减去这些数就好
1 #include <iostream> 2 #include <sstream> 3 #include <fstream> 4 #include <string> 5 #include <vector> 6 #include <deque> 7 #include <queue> 8 #include <stack> 9 #include <set> 10 #include <map> 11 #include <algorithm> 12 #include <functional> 13 #include <utility> 14 #include <bitset> 15 #include <cmath> 16 #include <cstdlib> 17 #include <ctime> 18 #include <cstdio> 19 #include <cstring> 20 #define FOR(i, a, b) for(int i = (a); i <= (b); i++) 21 #define RE(i, n) FOR(i, 1, n) 22 #define FORP(i, a, b) for(int i = (a); i >= (b); i--) 23 #define REP(i, n) for(int i = 0; i <(n); ++i) 24 #define SZ(x) ((int)(x).size ) 25 #define ALL(x) (x).begin(), (x.end()) 26 #define MSET(a, x) memset(a, x, sizeof(a)) 27 using namespace std; 28 29 30 typedef long long int ll; 31 typedef pair<int, int> P; 32 int read() { 33 int x=0,f=1; 34 char ch=getchar(); 35 while(ch<'0'||ch>'9') { 36 if(ch=='-')f=-1; 37 ch=getchar(); 38 } 39 while(ch>='0'&&ch<='9') { 40 x=x*10+ch-'0'; 41 ch=getchar(); 42 } 43 return x*f; 44 } 45 const double pi=3.14159265358979323846264338327950288L; 46 const double eps=1e-6; 47 const int mod = 1e9 + 7; 48 const int INF = 0x3f3f3f3f; 49 const int MAXN = 50; 50 const int xi[] = {0, 0, 1, -1}; 51 const int yi[] = {1, -1, 0, 0}; 52 53 int N, T; 54 int a[MAXN][MAXN], cnt, prime[MAXN]; 55 void init() { 56 for(int i = 2; i <= 51; i++) { 57 int flag= 1; 58 for(int j = 2; j*j <= i; j++) { 59 if(i%j == 0) { 60 flag = 0; 61 break; 62 } 63 } 64 if(flag) prime[i] = 1; 65 else prime[i] = 0; 66 } 67 memset(a, 0, sizeof(a)); 68 a[0][0] = 1; 69 for(int i = 1; i <= 25; i++) { 70 a[i][0] = 1; 71 for(int j = 1; j <= i; j++) { 72 a[i][j] = a[i-1][j] + a[i-1][j-1]; 73 } 74 75 } 76 } 77 int main() { 78 //freopen("in.txt", "r", stdin); 79 init(); 80 int k, s; 81 scanf("%d%d", &k, &s); 82 int sum = 0; 83 for(int i = 2; i <= s; i++) { 84 int n = s/i; 85 if(prime[i] && n >= k) { 86 sum += a[n][k]; 87 } 88 } 89 if(s/6>=k)sum -= a[s/6][k]; 90 if(s/10>=k) sum -= a[s/10][k]; 91 if(s/15>=k)sum -= a[s/15][k]; 92 if(s/14>=k)sum -= a[s/14][k]; 93 if(s/21>=k)sum -= a[s/21][k]; 94 if(s/22>=k) sum -= a[s/22][k]; 95 if(sum <= 10000) 96 printf("%d ", sum); 97 else printf("10000 "); 98 return 0; 99 }