1138 - Trailing Zeroes (III)
PDF (English) Statistics Forum
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
PROBLEM SETTER: JANE ALAM JAN
题意:给你一个数字,这个数字代表N!后面有几个0。给出这个数字,计算N的值。
分析:n!阶乘0的个数,就是求n!中5的重数,具体解释:https://blog.csdn.net/sdz20172133/article/details/81448221
这题加一个二分就行
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const double eps = 1e-8;
const ll MOD = 1e9 + 7;
const ll INFLL = 0x3f3f3f3f3f3f3f3f;
const int INF = 0x3f3f3f3f;
const int maxn = 1e5 + 10;
ll cal(ll n,ll p)
{
ll num=0;
while (n)
{
num += n/p;
n=n/ p;
}
return num;
}
int main()
{
int t;
int kase=0;
scanf("%d",&t);
while(t--)
{
kase++;
ll n;
scanf("%lld",&n);
ll l=1,r=1e18;
//cout<<cal(25,5)<<endl;
while(l<r)
{
ll mid=(l+r)>>1;
if(cal(mid,5)>=n)
{
r=mid;
}
else
{
l=mid+1;
}
}
printf("Case %d: ",kase);
if(cal(l,5)!=n)
{
printf("impossible\n");
}
else
printf("%lld\n",l);
}
return 0;
}