描述
Given an integer n, for all integers not larger than n, find the integer with the most divisors. If there is more than one integer with the same number of divisors, print the minimum one.
输入
One line with an integer n.
For 30% of the data, n ≤ 103
For 100% of the data, n ≤ 1016
输出
One line with an integer that is the answer.
#include<stdio.h>
#include<math.h>
int prime[13] = {2,3,5,7,11,13,17,19,23,29,31,37,41};
long long N, max = 1, ans = 1;
void DFS(long long now, long long divisor, int prime_index, long long pre){
if(max < divisor || (max == divisor && now < ans)){
max = divisor;
ans = now;
}
if(prime_index > 12) return;
long long i = 1, newnow;
while(i <= pre && now*pow(prime[prime_index], i) <= N){
newnow = now*pow(prime[prime_index], i);
DFS(newnow, divisor*(i+1), prime_index+1, i);
i++;
}
}
int main(){
scanf("%lld", &N);
DFS(1, 1, 0, log2(N));
printf("%lld\n", ans);
return 0;
}
本博客探讨了一个算法问题:给定一个整数 n,找出所有不大于 n 的整数中除数最多的整数,并在 n ≤ 10^16 的条件下给出解决方案。
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