最后两点RE，不知道问题出在哪里

#include <stdio.h>

int main(){
    int n;
    scanf("%d", &n);
    int m;
    scanf("%d", &m);
    int k, x = 1;
    for(k = 0; k<n-1; k++){
        x = x * 10;
    }

    //对数字串每位数字的提取方法如下：
    int a[n]; 
    for(k = 0; k<n; k++){
        int S = m / x;
        a[k] = S % 10;
        x = x / 10;
    }

    //用数表构建显示屏，1代表"X"，0代表"." 
    int b[5] [1005]; 
    int i;
    for(k = 0; k < 5; k++){
        for(i = 0; i < (4*n-1); i++){
            b[k][i] = 0;
        }
    }

    for(k = 0; k < n; k++){
        switch(a[k]){
            case 0:
                for(i=0; i<5; i++){
                    b[i][4*k] = 1;
                    b[i][4*k+2] = 1;

                } 
                b[0][4*k+1] = 1;
                b[4][4*k+1] = 1;
                break;
            case 1:
                for(i = 0; i < 5; i++){
                    b[i][4*k+2] = 1;
                }
                break;
            case 2:
                for(i = 4*k; i < (4*k +3); i++){
                    b[0][i] = 1;
                    b[2][i] = 1;
                    b[4][i] = 1;
                }
                b[1][4*k+2] = 1;
                b[3][4*k] = 1;
                break;
            case 3:
                for(i = 4*k; i < (4*k +3); i++){
                    b[0][i] = 1;
                    b[2][i] = 1;
                    b[4][i] = 1;
                }
                b[1][4*k+2] = 1;
                b[3][4*k+2] = 1;
                break;
            case 4:
                for(i = 0; i < 3; i++){
                    b[i][4*k] = 1;
                }
                for(i = 0; i < 5; i++){
                    b[i][4*k+2] = 1;
                }
                b[2][4*k+1] = 1;
                break;//每段case分支别忘加break 
            case 5:
                for(i = 4*k; i < (4*k +3); i++){
                    b[0][i] = 1;
                    b[2][i] = 1;
                    b[4][i] = 1;
                }
                b[1][4*k] = 1;
                b[3][4*k+2] = 1;
                break;
            case 6:
                for(i = 4*k; i < (4*k +3); i++){
                    b[0][i] = 1;
                    b[2][i] = 1;
                    b[4][i] = 1;
                }
                b[1][4*k] = 1;
                b[3][4*k+2] = 1;
                b[3][4*k] = 1;
                break;
            case 7:
                for(i = 0; i < 5; i++){
                    b[i][4*k+2] = 1;
                }
                b[0][4*k] = 1;
                b[0][4*k+1] = 1;
                break;  
            case 8:
                for(i = 4*k; i < (4*k +3); i++){
                    b[0][i] = 1;
                    b[2][i] = 1;
                    b[4][i] = 1;
                }
                b[1][4*k] = 1;
                b[3][4*k+2] = 1;
                b[3][4*k] = 1;
                b[1][4*k+2] = 1;
                break;
            default:
                for(i = 4*k; i < (4*k +3); i++){
                    b[0][i] = 1;
                    b[2][i] = 1;
                    b[4][i] = 1;
                }
                b[1][4*k] = 1;
                b[3][4*k+2] = 1;
                b[1][4*k+2] = 1;
        } 
    }

    //输出结果 
    for(k = 0; k < 5; k++){
        for(i = 0; i < (4*n-1); i++){
            if(b[k][i]){
                printf("X");
            } else{
                printf(".");
            }
        }
        printf("\n");
    }

    return 0;
}