【codechef】 Zeroes（大数灵活题）

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Virat loves mathematical questions. The other day he came across an interesting question which required him to find out the number of trailing zeroes for the
function. F(n) = 1¹*2²......N^N,where N is an integer. Virat solved the problem after a few attempts. He asked the same
question from his friend Rohit to see if he can solve it. Rohit couldn’t arrive at a solution and has asked for your help. Can you help him out?

Input

The first line contains a single integer T, the number of test cases. T test cases follow.
Each line contains a single integer
N.

Output

For each test case, output a single line containing a single integer which denotes the number of trailing zeroes for the value of the function computed for

Constraints

1 <= T <= 1000
1 <= N <= 10000

Example

Input:35102Output:5150
http://www.codechef.com/problems/ZEROES
F(n) = 1¹*2²......N^N，求F(n)的末位有几个0。这种题目就是要找规律。因为值实在是太大了，只能算出前几个，于是我就用了字符串大数模板，也只能勉强算出前200个，规律找得不够彻底。。。
0000 55555 1515151515（为方便我把5个数缩成1个）30 50 100 130 165 205 250 350 405 465 530 600 750 830 915 1005 1100 1300 1405 1515 1630 1750 2125....
我们知道 0 1 3 6 10 15....这个数列的前n项和为 n*(n+1)/2 ，上面白色背景的数字就是它的5倍，间隔为5（前面省去一样的了），红色部分减去白色的值是它的25倍，间隔为25，绿色部分减去红色的值是它的125倍，间隔为125，后面以此类推。
#include<iostream>  #include<algorithm>  #include<string>  #include<map>  #include<vector>  #include<cmath>  #include<string.h>  #include<stdlib.h>  #include<cstdio>  #define ll long long  using namespace std;int main(void){int t;scanf("%d",&t);while(t--){int n;cin>>n;ll s=0;for(int i=1;pow(5.0,i)<=n;++i){int m=n/pow(5.0,i);s+=pow(5.0,i)*(m+1)*m/2;}printf("%I64d\n",s);}return 0;}  

#include<iostream>  #include<algorithm>  #include<string>  #include<map>  #include<vector>  #include<cmath>  #include<string.h>  #include<stdlib.h>  #include<cstdio>  #define ll long long  using namespace std;int main (void){int t;scanf("%d",&t);while(t--){int n,i;long long int zero=0;scanf("%d",&n);for(i=5;i<=n;i+=5){if(i%78125 == 0)zero += i*7;else if(i%12625 == 0)zero += i*6;else if(i%3125 == 0)zero += i*5;else if(i%625 == 0)zero += i*4;else if(i%125 == 0)zero += i*3;else if(i%25 == 0)zero += i*2;elsezero += i;}printf("%lld\n",zero);}}

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