SGU[154] Factorial

DescriptionDescribe

You task is to find minimal natural number n, so, n! Contains exactly Q zeroes on the trail in decimal notation. As you know N! = 1*2*...*n. For example, 5! = contains one zero on the trail.

Your task is to find a minimum natural number n, so that the decimal below the n! end contains Q zeros. As you know, n! = 1 * 2 * ... * N. For example, 5! The end of = 120,120 contains 1 0.

InputInput

One number Q written in the input (0<=q<=10^8).

The input contains a number q (0 <= q <= 10^8).

OutputOutput

Write "No solution", if there is No such number N, and N otherwise.

If there is no such n, output "no solution", otherwise output n.

Sample InputSample input

2

Sample OutputSample output

10

AnalysisAnalysis

Statistics n! The number of the end 0, in fact, as long as the factor 2 and 5 number of the minimum value, because only 2 * 5 will produce 0. However, the number of factor 2 is much larger than the number of factor 5, so just look at the number of factor 5.

Because the space limit given by the topic is only 4096KB, so the table can not be played, will be MLE. Baidu after the finding can be used two points.

Two minutes when statistics 1 to n this n number of factor 5 number, we use this method: ans = n/5 + n/5^2 + n/5^3 + ...

There are two ways to deal with this process, one is to play the table, the power of 5 to play the table out, there is a similar Qin algorithm using the idea, each ans + = N/5, while N/= 5.

It is important to note that when you enter 0, you want to output 1, because 0 is not a natural number. The upper and lower bounds of the two points should be [1, 2147483647].

SolutionSolution Solutions

#include <iostream>using namespace Std;const int MAX = 2147483647;int Main () {int n;while (cin >> N) {if (n = = 0) {cout << 1 << endl; continue;} int L = 1, r = MAX, ans = -1;while (l <= r) {int nmid = (L + R) >> 1;int ntmp = nmid, ncnt = 0;while (ntmp) {ncnt + = NTMP/5; Ntmp/= 5; }if (ncnt = = N) {ans = Nmid; r = nMid-1;} else if (ncnt < N) {L = Nmid + 1;} else {r = nMid-1;}} if (ans = =-1) {cout << "No solution" << Endl;} else {cout << ans << Endl;}} return 0;}

　　

This topic began to WA several times, and later found that the upper bound of the two points is too small to set.

SGU[154] Factorial