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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 starts with an integer T (≤ 10000), denoting the number of test cases.
Each case contains an integer Q (1 ≤ Q ≤ 108) in a line.
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 |
题目描述：
假设有一个数n，它的阶乘末尾有Q个零，现在给出Q，问n最小为多少？
解题思路:
由于数字末尾的零等于min（因子2的个数，因子5的个数），又因为2<5,那么假设有一无限大的数n，n=2^x=5^y，可知x<<y。
所以我们可以直接根据因子5的个数，算阶乘末尾的零的个数。1<=Q<=10^8,所以可以用二分快速求解。
代码：
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 5 using namespace std; 6 7 long long slove(long long n) 8 { 9 long long ans = 0; 10 while(n) 11 { 12 ans += n/5; 13 n /= 5; 14 } 15 return ans; 16 } 17 18 int main() 19 { 20 int t, l = 1; 21 22 scanf("%d", &t); 23 while(t--) 24 { 25 long long n; 26 scanf("%lld", &n); 27 long long mid, low = 4, high = 500000000; 28 29 while(low <= high) // 二分查找快~ 30 { 31 mid = (low+high)/2; 32 long long num = slove(mid); 33 if(num >= n) 34 high = mid-1; 35 else 36 low = mid+1; 37 } 38 if(slove(low) == n) 39 printf("Case %d: %lld\n", l++, low); 40 else 41 printf("Case %d: impossible\n", l++); 42 } 43 return 0; 44 }
Trailing Zeroes (III) -；lightoj 1138
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原文地址：http://www.cnblogs.com/Tinamei/p/4746828.html
