2021.10.03

数据结构计算多组前进时间

#include

struct time
{
  int hours;
  int minutes;
  int seconds;  
};

struct time timeupdate(struct time now,int plus);

int main(void)
    {
        struct time testtime[5]={{11,59,59},{12,0,0},{1,29,59},{23,59,59},{19,12,27}};
        int i;
        int plus;
        printf("你想要时间向前多少秒？n");
        scanf("%d",&plus);
        for(i=0;i<5;i++)
            {
              printf("Time is %.2d:%.2d:%.2dn",testtime[i].hours,testtime[i].minutes,testtime[i].seconds);
          
          testtime[i]=timeupdate(testtime[i],plus);
          printf("%d seconds later time is %.2d:%.2d:%.2dn",plus,testtime[i].hours,testtime[i].minutes,testtime[i].seconds);
        
            }
        return 0;
    }
    
    
    struct time timeupdate(struct time now,int plus)
        {
            now.seconds+=plus;
            if(now.seconds>59)
                {
                    now.minutes=now.minutes+now.seconds/60;
                    
                    now.seconds=now.seconds%60;
                    
                    if(now.minutes>59)
                        {
                            now.hours+=now.minutes/60;
                            
                            now.minutes=now.minutes%60;
                            
                            if(now.hours>23)
                                {
                                    now.hours=now.hours%24;
                                }										
                            
                        }
                }
                return now;
        }

#include
int main()
{
	int n;
	int count = 0;
	int s,k; 
	int i,j;
	int b = 0;
	int a[10];
	int sum = 0;
	scanf("%d",&n);
	k = n;
	if(n==0)//单个0时输出
	{
		printf("a");
	}
	while(n!=0)//将数字分解，并存入数组
	{
		i = n % 10;
		a[count] = i;
		count++;
		n = n / 10;
	}
	s = count + 1;
	for(j = count-1;j>=0;j--)
	{
		s--;
		if(a[j]==0&&a[j-1]==0&&a[1]!=0&&k>10000&&b==0)//不超过亿，中间连续多0解决方法
		{
			b++;
			printf("W");
		}
		if(a[j]==0&&a[j-1]==0)//遇到连续0时的方法
		{
			continue;
		}
		switch(a[j])//数字转字母
		{
			case 0:
				printf("a");
				break;
			case 1:
				printf("b");
				break;
			case 2:
				printf("c");
				break;
			case 3:
				printf("d");
				break;
			case 4:
				printf("e");
				break;
			case 5:
				printf("f");
				break;
			case 6:
				printf("g");
				break;
			case 7:
				printf("h");
				break;
			case 8:
				printf("i");
				break;
			case 9:
				printf("j");
				break;
		}	
		
		if(a[j]!=0)//数字转单位
		{
			switch(s)
			{
				case 1:
					break;
				case 2:
					printf("S");
					break;
				case 3:
					printf("B");
					break;
				case 4:
					printf("Q");
					break;
				case 5:
					printf("W");
					break;
				case 6:
					printf("S");
					break;
				case 7:
					printf("B");
					break;
				case 8:
					printf("Q");
					break;
				case 9:
					printf("Y");
					break;
			}	
		}
		
		//s--;
	}
	return 0;
}