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 |
题意:求出现一个最小的数的阶乘满足末尾有连续q个0。
思路:一个最小的数末尾要出现0一定要是能被5或10整除的,0的连续长度与数组大小正相关,故二分数字得解。
#include<set>
#include<map>
#include<stack>
#include<cmath>
#include<queue>
#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<limits.h>
#include<algorithm>
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
const int INF=1e9;
typedef unsigned long long ll;
typedef long long LL;
int main()
{
int T,cases=;
scanf("%d",&T);
while(T--)
{
int q;
LL ans=-;
scanf("%d",&q);
int l=,r=q*,mid;//为什么右端点是q*5,因为q*5>=ans.想一想为什么(hint:刚才那个序列。。。)
while(l<=r)
{
mid=(l+r)/;
int m=mid,wu=,count=;
while(m>=wu)
{
int p=m;
p/=wu;
count+=p;
wu*=;
}
if(count==q)
{
ans=mid;
break;
}
else if(count>q)
r=mid-;
else
l=mid+;
}
if(ans==-)
printf("Case %d: impossible\n",cases++);
else
printf("Case %d: %lld\n",cases++,ans/*);
}
}
代码转自(https://blog.****.net/duan_1998/article/details/72566106)