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2018/7/31-zznu-oj-问题 B: N! 普拉斯 -【求大数的阶乘-ll存不下-然后取尾零的个数输出-暴力模拟】

程序员长春技术 2022年11月11日

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问题 B: N! 普拉斯

时间限制: 1 Sec 内存限制: 128 MB
提交: 114 解决: 35
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题目描述

在处理阶乘时也需要借助计算器。
在计算机中，数字是通过像01像素矩阵来显示的，最终的显示效果如下：

宝儿姐一直在思考一个问题，N!末尾究竟有多少个0？我们假设N!末尾有k个0，请按照规则打印k。

输入

输入一个正整数n（n< 50），输入以EOF结尾。

输出

我们假设N!末尾有k个0，请按照规则打印k，数字之间间隔3列0。

样例输入

样例输出

大致思路：

　　1、拿上一题的三维字符数组来用，定义也是一致的！

　　2、看清题意，尾零的个数不超过两位数，其实！可以直接特判输出！

　　3、要求输出时，数字之间间隔3列0，别忘了！

　　4、n不大，可以每次把尾零截出来，然后仅保留后十七位——或许不对吧！2333

题解：

 #include<iostream>
 #include<stdio.h>
 #include<string.h>
 #include<string>
 #include<vector>
 #include<algorithm>
 #define ll long long
 using namespace std;
 #define N 100 char s1[];
 int sum[];
 vector<int>order; char num[][][]={
      {"","","","","",
 "",""},
     {"","","","","",
 "",""},
 {"","","","","",
                     "",""},
 {"","","","","",
 "",""},
 {"","","","","",
 "",""},
  {"","","","","",
 "",""},
  {"","","","","",
 "",""},
 {"","","","","",
 "",""},
  {"","","","","",
 "",""},
  {"","","","","",
 "",""},
 {"","","","","",
 "",""}}; void paint( )   //绘制01字符串！
 {
     for(int i=;i<=;i++){
         for(int j=(int)order.size()-;j>=;j--){
             int x=order[j];
             printf("%s",num[x][i]);
             if(j>)
                 printf("%s",num[][i]);
         }
         cout<<endl;
     }
 } int cul(int n){
     int len=;
     sum[]=;
     for(int i=;i<=n;i++){
         for(int j=;j<len;j++){
             sum[j]*=i;
         }
         int j=;
         while(j<len){
             if(sum[j]>=){
                 sum[j+]+=sum[j]/;
                 sum[j]%=;
                 if(j+>=len)
                     len++;
             }
             j++;
         }
     }
     return len;
 } int main()
 {
     int n;
     while(scanf("%d",&n)!=EOF){         memset(sum,,sizeof(sum));
         int len3=cul(n);   //计算n的阶乘！        /* for(int i=0;i<len3;i++){
             printf("%d",sum[i]);
         }
         cout<<"*****"<<endl;
 */
         int i=;
         while(sum[i]==) //计算开头的零的数量!!
             i++;         order.clear();
         if(i==)
             order.push_back();
        while(i>){
             order.push_back(i%);
             i/=;
        }        paint();
     }     return ;
 }