TOJ1031

http://acm.tju.edu.cn/toj/showp1031.html

这道题参考：https://blog.csdn.net/aacm1992/article/details/7818652

之前在数字逻辑电路的课上学过7段数字显示的相关知识，可以应用到这道题。将每一个数字看作是由7段组成的，然后根据具体的数字对应起来哪一段“亮”。

此外，还学到了memset的相关知识：https://www.cnblogs.com/heyonggang/p/3419574.html

memset(MAP, 0, sizeof(MAP));

上面这句话代表将MAP数组的所有内容都设为ASCII码中的0（空字符），常用于内存空间初始化。

此外，这道题还需要注意的是'-'和'|'之间的位置关系、每个数字之间的位置关系。

代码：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

char MAP[200][200];

char Display(int s, int num, int pos){
    //左上“竖”
    if(num == 0 || num == 4 || num == 5 || num == 6 || num == 8 || num == 9){
        for(int i = 1; i < s+1; i++){
            MAP[i][pos] = '|';
        }
    }
    //上“横”
    if(num == 0 || num == 2 || num == 3 || num == 5 || num == 6 || num == 7 || num == 8 || num == 9){
        for(int j = pos+1; j < s+pos+1; j++){
           MAP[0][j] = '-';
        }
    }
    //右上“竖”
    if(num == 0 || num == 1 || num == 2 || num == 3 || num == 4 || num == 7 || num == 8 || num == 9){
        for(int i = 1; i < s+1; i++){
            MAP[i][pos+s+1] = '|';
        }
    }
    //中间“横”
    if(num == 2 || num == 3 || num == 4 || num == 5 || num == 6 || num == 8 || num == 9){
        for(int j = pos+1; j < s+pos+1; j++){
           MAP[s+1][j] = '-';
        }
    }
    //左下“竖”
    if(num == 0 || num == 2 || num == 6 || num == 8){
        for(int i = s+2; i < 2*s+2; i++){
            MAP[i][pos] = '|';
        }
    }
    //右下“竖”
    if(num == 0 || num == 1 || num == 3 || num == 4 || num == 5 || num == 6 || num == 7 || num == 8 || num == 9){
        for(int i = s+2; i < 2*s+2; i++){
            MAP[i][pos+s+1] = '|';
        }
    }
    //下“横”
    if(num == 0 || num == 2 || num == 3 || num == 5 || num == 6 || num == 8 || num == 9){
        for(int j = pos+1; j < s+pos+1; j++){
           MAP[2*s+2][j] = '-';
        }
    }
}

int main(){
    int s;
    char n[9];

    while(scanf("%d %s", &s, &n)){
        if(s == 0 && n[0] == '0' && n[1] == '\0'){
            break;
        }
        int len = strlen(n);
        //对二维数组初始化
        memset(MAP, 0, sizeof(MAP));
        for(int i = 0; i < 2*s+3; i++){
            for(int j = 0; j < len*(s+3)-1; j++){
                MAP[i][j] = ' ';
            }
        }
        for(int i = 0; i < len; i++){
            Display(s, n[i]-'0', (s+3)*i);
        }
        for(int i = 0; i < 2*s+3; i++){
            printf("%s\n", MAP[i]);
        }
        printf("\n");
    }
    return 0;
}