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
2
样例输出
01110 10001 10011 10101 11001 10001 01110
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 }
（暴力模拟题，一定要耐住性子，不急不躁，稳住才能AC）
原文地址：https://www.cnblogs.com/zhazhaacmer/p/9399455.html
时间： 2024-10-07 20:42:21