1091. Tmutarakan ExamsTime limit: 1.0 second 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. InputThe input contains numbers K and S (2 ≤ K ≤ S ≤ 50). OutputYou 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
Problem Author: Stanislav Vasilyev Tags: number theory (hide tags for unsolved problems) Difficulty: 480 Printable version Submit solution Discussion (29) |
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思路：和http://www.cnblogs.com/zzuli2sjy/p/5467008.html一样；
1 #include<stdio.h>
2 #include<string.h>
3 #include<iostream>
4 #include<algorithm>
5 #include<queue>
6 using namespace std;
7 typedef long long LL;
8 bool prime[100];
9 int ans[100];
10 int coutt[10000];
11 LL dp[60][60];
12 int ask[100];
13 int id[100];
14 queue<int>que;
15 int main(void)
16 {
17 int i,j,k,p,q;
18 dp[0][0]=1;
19 dp[1][0]=1;
20 dp[1][1]=1;
21 for(i=2; i<=60; i++)
22 {
23 for(j=0; j<=60; j++)
24 {
25 if(j==0||i==j)
26 {
27 dp[i][j]=1;
28 }
29 else dp[i][j]=dp[i-1][j-1]+dp[i-1][j];
30 }
31 }
32 for(i=2; i<=10; i++)
33 {
34 if(!prime[i])
35 {
36 for(j=i; i*j<=50; j++)
37 {
38 prime[i*j]=true;
39 }
40 }
41 }
42 int cnt=0;
43 for(i=2; i<=50; i++)
44 {
45 if(!prime[i])
46 {
47 ans[cnt++]=i;
48 }
49 }
50 while(scanf("%d %d",&p,&q)!=EOF)
51 { int s;
52 memset(coutt,0,sizeof(coutt));
53 for(s=2; s<=q; s++)
54 {
55 int cc=s;
56 int flag=0;
57 int t=0;
58 while(cc>1)
59 {
60 if(cc%ans[t]==0&&flag==0)
61 {
62 flag=1;
63 que.push(ans[t]);
64 cc/=ans[t];
65 }
66 else if(cc%ans[t]==0)
67 {
68 cc/=ans[t];
69 }
70 else
71 {
72 t++;
73 flag=0;
74 }
75 }
76 int vv=0;
77 while(!que.empty())
78 {
79 ask[vv++]=que.front();
80 que.pop();
81 }
82 for(i=1; i<=(1<<vv)-1; i++)
83 {
84 LL sum=1;
85 int dd=0;
86 for(j=0; j<vv; j++)
87 {
88 if(i&(1<<j))
89 {
90 dd++;
91 sum*=ask[j];
92 }
93 }
94 id[sum]=dd;
95 coutt[sum]++;
96
97 }
98 }
99 LL summ=0;
100 for(i=2; i<=50; i++)
101 {
102 if(id[i]%2&&coutt[i]>=p)
103 {
104 summ+=dp[coutt[i]][p];
105 }
106 else if(coutt[i]>=p)summ-=dp[coutt[i]][p];
107 }if(summ>=10000)summ=10000;
108 printf("%lld\n",summ);
109 }
110 return 0;
111 }