wenbao与反素数
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对于任何正整数x,其约数的个数记做g(x).例如g(1)=1,g(6)=4.如果某个正整数x满足:对于任意i(0<i<x),都有g(i)<g(x),则称x为反素数·
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1 #include <iostream> 2 #include <string.h> 3 #include <algorithm> 4 using namespace std; 5 #define LL long long 6 const int N = 500010; 7 const int M = 1024; 8 9 struct node { 10 int a, b; 11 node() {} 12 node(int x, int y) : a(x), b(y) {} 13 bool operator < (const node cmp) const { 14 return b < cmp.b; 15 } 16 }f[M]; 17 18 bool vis[M]; 19 int prime[M]; 20 int cnt, num; 21 22 void get_prime() { 23 int i, j; 24 memset(vis, true, sizeof(vis)); 25 for(i = 2; i < M; ++i) { 26 for(j = i*i; j < M; j += i) { 27 vis[j] = false; 28 } 29 } 30 cnt = 0; 31 for(i = 2; i < M; ++i) { 32 if(vis[i]) prime[cnt++] = i; 33 } 34 } 35 36 void solve(int pnum, int val, int pos, int lim) { 37 if(val > N) return ; 38 f[num++] = node(pnum, val); 39 LL nlim, nval, npnum; 40 nval = val; nlim = 0; npnum = 1; 41 while(nlim < lim && nval <= N) { 42 nlim++; npnum++; nval *= prime[pos]; 43 if(nval <= N) solve(npnum*pnum, nval, pos + 1, nlim); 44 } 45 } 46 void get_antiprime() { 47 get_prime(); 48 num = 0; 49 solve(1, 1, 0, 32); 50 sort(f, f + num); 51 for(int i = 0; i < num; ++i){ 52 printf("%d--%d\n", f[i].a, f[i].b); 53 } 54 } 55 int main(){ 56 get_antiprime(); 57 return 0; 58 } 59 60 61 const int antiprime[]={1,2,4,6,12,24,36,48,60,120,180,240,360,720,840, 62 1260,1680,2520,5040,7560,10080,15120,20160,25200, 63 27720,45360,50400,55440,83160,110880,166320,221760, 64 277200,332640,498960,554400,665280 65 }; 66 67 const int factorNum[]={1,2,3,4,6,8,9,10,12,16,18,20,24,30,32,36,40,48,60, 68 64,72,80,84,90,96,100,108,120,128,144,160,168,180, 69 192,200,216,224 70 };
1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 11 11 12 12 13 13 14 14 15 15 16 16 17 17 18 18 19 19 20 20 21 21 22 22 23 23 24 24 25 25 26 26 27 27 28 28 29 29 30 30 31 31 #include<iostream> 32 #include<cstdio> 33 #include<cstring> 34 #define inf 0x7fffffff 35 #define ll long long 36 using namespace std; 37 int n,ans=1,num=1; 38 int p[15]={1,2,3,5,7,11,13,17,19,23,29,31}; 39 void dfs(int k,ll now,int cnt,int last) 40 { 41 if(k==12) 42 { 43 if(now>ans&&cnt>num){ans=now;num=cnt;} 44 if(now<=ans&&cnt>=num){ans=now;num=cnt;} 45 return; 46 } 47 int t=1; 48 for(int i=0;i<=last;i++) 49 { 50 dfs(k+1,now*t,cnt*(i+1),i); 51 t*=p[k]; 52 if(now*t>n)break; 53 } 54 } 55 int main() 56 { 57 scanf("%d",&n); 58 dfs(1,1,1,20); 59 printf("%d",ans); 60 return 0; 61 }
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