2019ECNU机试06

Code

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
#pragma once
#define _CRT_SECURE_NO_WARNINGS
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

//这里这个计算方法会存在误差,当找到mid需要进一步的缩小值的大小
int cnt(int n) {
int base = 5,sum=0;
while (n/base!=0) {
sum += n / base;
base *= 5;
}
return sum;
}

int fun() {
int k,n;
scanf("%d", &k);

int mid, temp, begin=1, end=2e8,latest;

while (begin <= end) {
mid = begin+(end-begin)/2;
temp = cnt(mid);
if (temp == k) {
latest = mid;
break;
}

if (temp>k) {
latest = mid;
//说明在左边
end = mid - 1;
}
else if(temp<k){
begin = mid + 1;
}
}
//进一步缩小值的范围
while (cnt(latest-1)>=k)
latest--;

printf("%d\n", latest);

return 0;
}

----\(˙<>˙)/----赞赏一下吧~