ZOJ2562
https://vjudge.net/problem/11781/origin
<=n的且因子数最多的那个数
做法:同因子数取最小,dfs更新答案
#include <iostream> #include <cstdio> #include <queue> #include <algorithm> #include <cmath> #include <cstring> #define inf ~0 #define N 1000010 #define p(a) putchar(a) #define For(i,a,b) for(unsigned long long i=a;i<=b;++i) using namespace std; unsigned long long n,ans,now; unsigned long long prime[]={2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53}; void in(unsigned long long &x){ unsigned long long y=1;char c=getchar();x=0; while(c<‘0‘||c>‘9‘){if(c==‘-‘)y=-1;c=getchar();} while(c<=‘9‘&&c>=‘0‘){ x=(x<<1)+(x<<3)+c-‘0‘;c=getchar();} x*=y; } void o(unsigned long long x){ if(x<0){p(‘-‘);x=-x;} if(x>9)o(x/10); p(x%10+‘0‘); } void dfs(unsigned long long depth,unsigned long long num,unsigned long long cnt,unsigned long long up){ if(num>=now&&cnt<=ans) return; if(ans<cnt){ ans=cnt; now=num; } if(ans==cnt&&now>num) now=num; For(i,1,up){ if(num*prime[depth]>n) return; dfs(depth+1,num*=prime[depth],cnt*(i+1),i); } } int main(){ while(scanf("%llu",&n)!=EOF){ ans=0; now=0; dfs(0,1,1,64); o(now); p(‘\n‘); } return 0; }
原文:https://www.cnblogs.com/war1111/p/11297213.html