Factorial
Problem Description
You task is to find minimal natural number N, so that N! contains exactly Q zeroes on the trail in decimal notation. As you know N! = 12…*N. For example, 5! = 120, 120 contains one zero on the trail.
Input
One number Q written in the input (0<=Q<=10^8).
Output
Write “No solution”, if there is no such number N, and N otherwise.
Sample test(s)
Input
2
Output
10
Solution
代码语言：javascript复制#include <bits/stdc .h>
int main() {
std::ios::sync_with_stdio(false);
int q;
std::cin >> q;
if(q == 0) {
std::cout << 1 << std::endl;
}
long long left = 5;
long long right = 5 * q;
long long res = -1;
while(left <= right) {
long long mid = (left right) / 2;
long long tmpMid = mid;
int count = 0;
while(tmpMid != 0) {
tmpMid /= 5;
count = tmpMid;
}
if(count == q) {
res = mid;
right = mid - 1;
} else if(count > q) {
right = mid - 1;
} else {
left = mid 1;
}
}
if(res == -1) {
std::cout << "No solution" << std::endl;
} else {
std::cout << res << std::endl;
}
return 0;
}
