几道略有难度的C语言题 求高人给出源代码解决办法

#include<stdio.h>

/*使用结构体来表示大楼*/
struct building {
    int length;
    int width;
    int height;
};

int main(){

    /*记录多少组数据*/
    int group;
    /*记录每一组数据里面有多少个大楼，我假定的最多50，这个可以是N*/
    int count[50];
    int i, j, k;
    /*记录总共有多少个大楼*/
    int temp = 0;
    /*记录输入的大楼，假定最多100个大楼*/
    struct building building[100];
    /*中间变量，排序时使用*/
    struct building tempBuilding;

    printf("Please enter the group of data:");
    scanf("%d", &group);
    for(i=0;i<group;i++){
        printf("Please enter the count of building:");
        scanf("%d", &count[i]);
        for(j=0; j< count[i]; j++){
            printf("Please enter the length width height of this building:");
            scanf("%d%d%d", &building[temp].length, &building[temp].width, &building[temp].height);
            temp ++;
        }
    }

    /*对每一组的大楼进行排序，时间复杂度不够好，不过对于这题的要求是没有关系的，凑合着使吧*/
    for(i=0;i<group;i++){
        for(j=i*group;j<(count[i]-1)*group;j++){
            for(k=j;k<(count[i]-1)*group;k++){
                if(building[j].height < building[k+1].height){
                    tempBuilding = building[k+1];
                    building[k+1] = building[j];
                    building[j] = tempBuilding;
                }
                else if(building[j].height == building[k+1].height && building[j].length*building[j].width < building[k+1].length*building[k+1].width){
                    tempBuilding = building[k+1];
                    building[k+1] = building[j];
                    building[j] = tempBuilding;
                }
            }
        }
    }

    /*输出排序后的大楼*/
    for(i=0;i<temp;i++){
        printf("%d\t%d\t%d \n", building[i].length, building[i].width, building[i].height);
    }


    getch();
    return 0;
}