uva706 LC-Display
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LC-Display
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 file contains several lines, one for each number to be displayed. Each line contains two integerss, n ( ), 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.
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.)
把数分5部分 横切 每次打印一个水平条
#include <stdio.h>#include <string.h>#define max 50 char a1[] = "- -- -----";char a2_1[] = "| ||| ||";char a2_2[] = "||||| |||";char a3[] = "- ----- --";char a4_1[] = "| | | | ";char a4_2[] = "|| |||||||";char a5[] = "- -- -- --";int main(){ int s; char n[max]; while (scanf("%d %s", &s, n) && (s || n[0] - '0')) { int l = strlen(n); for (int i = 0; i < l; i++) { printf(" "); int num = n[i] - '0'; for (int j = 0; j < s; j++) { printf("%c", a1[num]); } printf(" "); } printf("\n"); for (int i = 0; i < s; i++) { for (int j = 0; j < l; j++) { int num = n[j] - '0'; printf("%c", a2_1[num]); for (int k = 0; k < s; k++) printf(" "); printf("%c", a2_2[num]); printf(" "); } printf("\n"); } for (int i = 0; i < l; i++) { printf(" "); int num = n[i] - '0'; for (int j = 0; j < s; j++) { printf("%c", a3[num]); } printf(" "); } printf("\n"); for (int i = 0; i < s; i++) { for (int j = 0; j < l; j++) { int num = n[j] - '0'; printf("%c", a4_1[num]); for (int k = 0; k < s; k++) printf(" "); printf("%c", a4_2[num]); printf(" "); } printf("\n"); } for (int i = 0; i < l; i++) { printf(" "); int num = n[i] - '0'; for (int j = 0; j < s; j++) { printf("%c", a5[num]); } printf(" "); } printf("\n"); }}
LC-Display 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 file contains several lines, one for each number to be displayed. Each line contains two integerss, n ( ), 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.
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.)
把数分5部分 横切 每次打印一个水平条
#include <stdio.h>#include <string.h>#define max 50 char a1[] = "- -- -----";char a2_1[] = "| ||| ||";char a2_2[] = "||||| |||";char a3[] = "- ----- --";char a4_1[] = "| | | | ";char a4_2[] = "|| |||||||";char a5[] = "- -- -- --";int main(){ int s; char n[max]; while (scanf("%d %s", &s, n) && (s || n[0] - '0')) { int l = strlen(n); for (int i = 0; i < l; i++) { printf(" "); int num = n[i] - '0'; for (int j = 0; j < s; j++) { printf("%c", a1[num]); } printf(" "); } printf("\n"); for (int i = 0; i < s; i++) { for (int j = 0; j < l; j++) { int num = n[j] - '0'; printf("%c", a2_1[num]); for (int k = 0; k < s; k++) printf(" "); printf("%c", a2_2[num]); printf(" "); } printf("\n"); } for (int i = 0; i < l; i++) { printf(" "); int num = n[i] - '0'; for (int j = 0; j < s; j++) { printf("%c", a3[num]); } printf(" "); } printf("\n"); for (int i = 0; i < s; i++) { for (int j = 0; j < l; j++) { int num = n[j] - '0'; printf("%c", a4_1[num]); for (int k = 0; k < s; k++) printf(" "); printf("%c", a4_2[num]); printf(" "); } printf("\n"); } for (int i = 0; i < l; i++) { printf(" "); int num = n[i] - '0'; for (int j = 0; j < s; j++) { printf("%c", a5[num]); } printf(" "); } printf("\n"); }}
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