POJ 1102 LC-Display 模拟
来源:互联网 发布:论文格式 知乎 编辑:程序博客网 时间:2024/04/29 14:46
Description
A friend of you has just bought a new computer. Until now, the most powerful computer he ever used has been a pocket calculator. Now, looking at his new computer, he is a bit disappointed, because he liked the LC-display of his calculator so much. So you decide to write a program that displays numbers in an LC-display-like style on his computer.
Input
The input contains several lines, one for each number to be displayed. Each line contains two integers s, n (1 <= s <= 10, 0 <= n <= 99 999 999), where n is the number to be displayed and s is the size in which it shall be displayed.
The input file will be terminated by a line containing two zeros. This line should not be processed.
The input file will be terminated by a line containing two zeros. This line should not be processed.
Output
Output the numbers given in the input file in an LC-display-style using s "-" signs for the horizontal segments and s "|" signs for the vertical ones. Each digit occupies exactly s+2 columns and 2s+3 rows. (Be sure to fill all the white space occupied by the digits with blanks, also for the last digit.) There has to be exactly one column of blanks between two digits.
Output a blank line after each number. (You will find a sample of each digit in the sample output.)
Output a blank line after each number. (You will find a sample of each digit in the sample output.)
Sample Input
2 123453 678900 0
Sample Output
-- -- -- | | | | | | | | | | | | -- -- -- -- | | | | | | | | | | -- -- -- --- --- --- --- --- | | | | | | | || | | | | | | || | | | | | | | --- --- --- | | | | | | | || | | | | | | || | | | | | | | --- --- --- ---输入的两个数分别是数字的尺寸 和要显示的数字组合 (注意:每两组数据空一行,每两个数字空一格)代码:#include<iostream>#include<cstdio>#include<cstring>using namespace std;int main(){ char e[20][3]= {" -","-"}; char q[33]; char w[1][2]= {"|"}; int m; int i,j,k; int t=0; while(scanf("%d%s",&m,q)!=EOF) { if(m==0) continue; if(t!=0) cout<<endl; t++; if(m==0&&q[0]=='0') break; for(i=0; i<=strlen(q)-1; i++) { if(i!=0) cout<<' '; if(q[i]=='1'||q[i]=='4') { for(j=1; j<=m+2; j++) cout<<' '; } else { cout<<' '; for(j=1; j<=m; j++) { cout<<e[1]; } cout<<' '; } } cout<<endl; for(i=1; i<=m; i++) { for(k=0; k<=strlen(q)-1; k++) { if(k!=0) cout<<' '; if(q[k]=='5'||q[k]=='6') { cout<<w[0]; for(j=0; j<=m; j++) cout<<' '; } else if(q[k]=='1'||q[k]=='2'||q[k]=='3'||q[k]=='7') { for(j=0; j<=m; j++) cout<<' '; cout<<w[0]; } else { cout<<w[0]; for(j=0; j<m; j++) cout<<' '; cout<<w[0]; } } cout<<endl; } for(i=0; i<strlen(q); i++) { if(i!=0) cout<<' '; if(q[i]=='1'||q[i]=='7'||q[i]=='0') { for(j=1; j<=m+2; j++) cout<<' '; } else { cout<<' '; for(j=1; j<=m; j++) { cout<<e[1]; } cout<<' '; } } cout<<endl; for(i=1; i<=m; i++) { for(k=0; k<=strlen(q)-1; k++) { if(k!=0) cout<<' '; if(q[k]=='2') { cout<<w[0]; for(j=0; j<=m; j++) cout<<' '; } else if(q[k]=='0'||q[k]=='8'||q[k]=='6') { cout<<w[0]; for(j=0; j<m; j++) cout<<' '; cout<<w[0]; } else { for(j=0; j<=m; j++) cout<<' '; cout<<w[0]; } } cout<<endl; } for(i=0; i<strlen(q); i++) { if(i!=0) cout<<' '; if(q[i]=='1'||q[i]=='7'||q[i]=='4') { for(j=1; j<=m+2; j++) cout<<' '; } else { cout<<' '; for(j=1; j<=m; j++) { cout<<e[1]; } cout<<' '; } } cout<<endl; } return 0;}
把没组数字分为五段 (三横二竖) 其实写出前两段 后三段也差不多了。。。
这次终于把速度提高了一点= =
显示出来的数字还挺 好看的 哈
0 0
- POJ 1102 LC-Display 模拟
- poj 1102 LC-Display(模拟)
- POJ 1102 LC-Display(模拟题)
- POJ 1102 LC DISPLAY
- poj 1102 LC-Display
- POJ 1102 LC-Display
- poj 1102 LC-Display
- POJ 1102 - LC-Display
- poj 1102 LC—display
- POJ 1102 LC Display 笔记
- 模拟 UVa 706 LC-Display
- 1102 LC-Display
- LC-Display
- LC-Display
- LC-Display
- LC-Display
- LC-Display
- LC-Display
- FTP协议
- 不同分辨率下,字体大小怎么适应?
- 定位获取位置及位置编码-反编码
- Binder机制5--- Binder实现进程管理服务示例
- 移植2.6.30.4到MINI2440
- POJ 1102 LC-Display 模拟
- ActiveMQ Master Slave配置以及示例
- 卸载mysql的方法
- 前端对接BO相关流程总结
- HDOJ 题目1992 Tiling a Grid With Dominoes(递推)
- 修改android theme的问题
- LeetCode-Insertion Sort List
- 量化研究: Julia还是Matlab?
- 关于插入排序