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】

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
[提交] [状态] [讨论版] [命题人:admin]

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：

  1 #include<iostream>
  2 #include<stdio.h>
  3 #include<string.h>
  4 #include<string>
  5 #include<vector>
  6 #include<algorithm>
  7 #define ll long long
  8 using namespace std;
  9 #define N 100
 10 
 11 char s1[100];
 12 int sum[100];
 13 vector<int>order;
 14 
 15 char num[12][10][10]={
 16      {"01110","10001","10011","10101","11001",
 17 "10001","01110"},
 18     {"00100","01100","00100","00100","00100",
 19 "00100","01110"},
 20 {"01110","10001","00001","00110","01000",
 21                     "10000","11111"},
 22 {"11111","00001","00010","00110","00001",
 23 "10001","01110"},
 24 {"00010","00110","01010","10010","11111",
 25 "00010","00010"},
 26  {"11111","10000","10000","11110","00001",
 27 "00001","11110"},
 28  {"01111","10000","10000","11110","10001",
 29 "10001","01110"},
 30 {"11111","00001","00010","00100","00100",
 31 "00100","00100"},
 32  {"01110","10001","10001","01110","10001",
 33 "10001","01110"},
 34  {"01110","10001","10001","01111","00001",
 35 "00010","11100"},
 36 {"000","000","000","000","000",
 37 "000","000"}};
 38 
 39 void paint( )   //绘制01字符串！
 40 {
 41     for(int i=0;i<=6;i++){
 42         for(int j=(int)order.size()-1;j>=0;j--){
 43             int x=order[j];
 44             printf("%s",num[x][i]);
 45             if(j>0)
 46                 printf("%s",num[10][i]);
 47         }
 48         cout<<endl;
 49     }
 50 }
 51 
 52 int cul(int n){
 53     int len=1;
 54     sum[0]=1;
 55     for(int i=1;i<=n;i++){
 56         for(int j=0;j<len;j++){
 57             sum[j]*=i;
 58         }
 59         int j=0;
 60         while(j<len){
 61             if(sum[j]>=10){
 62                 sum[j+1]+=sum[j]/10;
 63                 sum[j]%=10;
 64                 if(j+1>=len)
 65                     len++;
 66             }
 67             j++;
 68         }
 69     }
 70     return len;
 71 }
 72 
 73 int main()
 74 {
 75     int n;
 76     while(scanf("%d",&n)!=EOF){
 77 
 78         memset(sum,0,sizeof(sum));
 79         int len3=cul(n);   //计算n的阶乘！
 80 
 81        /* for(int i=0;i<len3;i++){
 82             printf("%d",sum[i]);
 83         }
 84         cout<<"*****"<<endl;
 85 */
 86         int i=0;
 87         while(sum[i]==0) //计算开头的零的数量!!
 88             i++;
 89 
 90         order.clear();
 91         if(i==0)
 92             order.push_back(0);
 93        while(i>0){
 94             order.push_back(i%10);
 95             i/=10;
 96        }
 97 
 98        paint();
 99     }
100 
101 
102     return 0;
103 }

View Code（暴力模拟题，一定要耐住性子，不急不躁，稳住才能AC）