"BZOJ3629" "JLOI2014" smart Stef DFS Prime sieve

Link:

#include <stdio.h>int main(){    puts("转载请注明出处[vmurder]谢谢");    puts("网址：blog.csdn.net/vmurder/article/details/44698555");}

Exercises

We find that by decomposing a number into a factorization, we can then calculate the approximate number of these numbers according to the number of each mass factor. So we can violently dismantle each number, the root-square time complexity decomposition.
is to enumerate how much it is used for each prime number, and then the count is removed and the next layer is deep.

Code:

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#define N 50100using namespace STD;intprime[n],cnt;BOOLVis[n];voidShakeintN) {intI,j,k; for(i=2; i<=n;i++) {if(!vis[i]) prime[++cnt]=i; for(j=1; j<=cnt&&i*prime[j]<=n;j++) {vis[i*prime[j]]=1;if(i%prime[j]==0) Break; }    }}BOOLCheckintx) {BOOLflag=0; for(intI=1;(Long Long) prime[i]*prime[i]<=x;i++)if(x%prime[i]==0)return 0;return 1;}intAns[n],num;voidDfsLong LongNowintRemain,intT) {if(remain==1) {Ans[++num]=now;return;}if(remain==2)return;if(remain-1>=prime[t]&&check (remain-1)) ans[++num]=now* (remain-1); for(intI=t; (Long Long) {prime[i]*prime[i]<=remain;i++) {Long Longsum=prime[i]+1, X=prime[i]; while(Sum<=remain) {if(remain%sum==0) DFS (now*x,remain/sum,i+1);        X*=prime[i],sum+=x; }    }}intNintMain () {//Freopen ("Test.in", "R", stdin);Shake50000); while(scanf("%d", &n)!=eof) {num=0; Dfs1N1); Sort (ans+1, ans+num+1), n=0; for(intI=1; i<=num;i++)if(ans[i]!=ans[i-1]) ans[++n]=ans[i];printf("%d\n", n);if(n>=1)printf("%d", ans[1]); for(intI=2; i<=n;i++)printf("%d", Ans[i]);if(num)puts(""); }return 0;}

"BZOJ3629" "JLOI2014" smart Stef DFS Prime sieve