【Light-oj】-1138-Trailing Zeroes (III)(找到一个数使其阶乘后0的个数为n)

1138 - Trailing Zeroes (III)

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Time Limit: 2 second(s)

Memory Limit: 32 MB

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! = 1*2*...*N. For example, 5! = 120, 120 contains one zero
on the trail.

Input

Input starts with an integer T (≤ 10000), denoting the number of test cases.

Each case contains an integer Q (1 ≤ Q ≤ 108) in a line.

Output

For each case, print the case number and N. If no solution is found then print 'impossible'.

Sample Input	Output for Sample Input
3 1 2 5	Case 1: 5 Case 2: 10 Case 3: impossible

1:先写一个函数，计算一个数阶乘后会有几个0

2:left和right分别定义为最小的0和最大的正整数，然后两边逼近找到一个数阶乘后0的个数为n。

#include<cstdio>
#define MAX 0x3f3f3f3f //不会超出的最好用的最大正整数
int gs(int a) //写一个函数计算出一个数的阶乘有几个0
{
int ant=0;
while(a)
{
ant+=a/5;
a=a/5;
}
return ant;
}
int main()
{
int t,num=1;
scanf("%d",&t);
while(t--)
{
int n,l=0,r=MAX,mid; //把left定义为最小0,right定义为最大
scanf("%d",&n);
printf("Case %d: ",num);
num++;
while(l<=r)
{
mid=(l+r)>>1;
if(gs(mid)>=n) //个数多于n则right左移换为mid-1
r=mid-1;
else
l=mid+1;
}
if(gs(l)==n)
printf("%d\n",l);
else
printf("impossible\n");
}
return 0;
}