链接地址：
Openjudge：http://bailian.openjudge.cn/practice/1102
Poj:http://poj.org/problem?id=1102
题目：
- 总时间限制:
- 1000ms
- 内存限制:
- 65536kB
- 描述
- 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.
- 输入
- 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.- 输出
- 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.)- 样例输入
- 样例输出
-- -- -- | | | | | | | | | | | | -- -- -- -- | | | | | | | | | | -- -- -- --- --- --- --- --- | | | | | | | | | | | | | | | | | | | | | | | | --- --- --- | | | | | | | | | | | | | | | | | | | | | | | | --- --- --- ---- 来源
思路：
模拟题，使用bieset保存结果，用iterator迭代数组，注意数字间有空格，每一行之间有空行
代码：
1 #include <iostream> 2 #include <bitset> 3 #include <string> 4 #include <sstream> 5 using namespace std; 6 7 8 int main() 9 { 10 string str0 = "1110111"; 11 string str1 = "0100100"; 12 string str2 = "1011101"; 13 string str3 = "1101101"; 14 string str4 = "0101110"; 15 string str5 = "1101011"; 16 string str6 = "1111011"; 17 string str7 = "0100101"; 18 string str8 = "1111111"; 19 string str9 = "1101111"; 20 21 string str = str9 + str8 + str7 + str6 + str5 + str4 + str3 + str2 + str1 + str0; 22 23 bitset<70> bs(str); 24 25 int s,n; 26 cin>>s>>n; 27 int flag; 28 string::iterator iter; 29 int i; 30 while(s!=0 || n != 0) 31 { 32 ostringstream s1; 33 s1<<n; 34 string str_n = s1.str(); 35 36 flag = 0; 37 for(iter = str_n.begin(); iter != str_n.end(); ++iter) 38 { 39 if(flag) cout<<" "; 40 else flag = 1; 41 int num = (*iter) - '0'; 42 cout<<" "; 43 for(int i = 0; i < s; i++) 44 { 45 if(bs[num * 7]) cout<<"-"; 46 else cout<<" "; 47 } 48 cout<<" "; 49 } 50 cout<<endl; 51 52 53 for(i = 0; i < s; i++) 54 { 55 flag = 0; 56 for(iter = str_n.begin(); iter != str_n.end(); ++iter) 57 { 58 if(flag) cout<<" "; 59 else flag = 1; 60 int num = (*iter) - '0'; 61 if(bs[num * 7 + 1]) cout<<"|"; 62 else cout<<" "; 63 for(int j = 0; j < s; j++) cout<<" "; 64 if(bs[num * 7 + 2]) cout<<"|"; 65 else cout<<" "; 66 } 67 cout<<endl; 68 } 69 70 flag = 0; 71 for(iter = str_n.begin(); iter != str_n.end(); ++iter) 72 { 73 if(flag) cout<<" "; 74 else flag = 1; 75 int num = (*iter) - '0'; 76 //step1 77 cout<<" "; 78 for(i = 0; i < s; i++) 79 { 80 if(bs[num * 7 + 3]) cout<<"-"; 81 else cout<<" "; 82 } 83 cout<<" "; 84 } 85 cout<<endl; 86 87 for(i = 0; i < s; i++) 88 { 89 flag = 0; 90 for(iter = str_n.begin(); iter != str_n.end(); ++iter) 91 { 92 if(flag) cout<<" "; 93 else flag = 1; 94 int num = (*iter) - '0'; 95 if(bs[num * 7 + 4]) cout<<"|"; 96 else cout<<" "; 97 for(int j = 0; j < s; j++) cout<<" "; 98 if(bs[num * 7 + 5]) cout<<"|"; 99 else cout<<" "; 100 } 101 cout<<endl; 102 } 103 104 flag = 0; 105 for(string::iterator iter = str_n.begin(); iter != str_n.end(); ++iter) 106 { 107 if(flag) cout<<" "; 108 else flag = 1; 109 int num = (*iter) - '0'; 110 //step1 111 cout<<" "; 112 for(i = 0; i < s; i++) 113 { 114 if(bs[num * 7 + 6]) cout<<"-"; 115 else cout<<" "; 116 } 117 cout<<" "; 118 } 119 cout<<endl;/**/ 120 cout<<endl; 121 cin>>s>>n; 122 } 123 return 0; 124 }