zn的手环

题目链接(- ->点击题目链接自行查看)

解题思路如下：
1.如果均为上午或下午，且起床时间比入睡时间晚，则可以直接相减
  否则起床时间 + 24再相减。
2.因为数字未0~9(无13、14...，本人就是看错了...)
  此时起床时间 + 12和入睡时间相减即可。

贴一份超时代码1001ms。。。。。。，全超时，我也是醉了~~

#include <iostream>
#include <cstring>
using namespace std;

int  starthour, startminute, endhour, endminute;
int  hour, minute;
char chstart, chend;

void Large()
{
	endminute += 60;
	--endhour;
}

int TimeProblemAAPPAP(int H, int M)
{
	if(startminute > endminute) Large();
	H = endhour - starthour;
	M = endminute - startminute;
	if(H == hour && M == minute) printf("YES\n");
	else
	{
		printf("NO\n");
		printf("%dh%dmin", H, M);
	}
}

int TimeProblemPA(int H, int M)
{
	H = 11 - starthour;
	M = 60 - startminute;
	H += endhour;
	M += endminute;
	if(M >= 60) 
	{
		M -= 60;
		++H;
	}
	if(H == hour && M == minute) printf("YES\n");
	else
	{
		printf("NO\n");
		printf("%dh%dmin", H, M);
	}
	
}


int main()
{
	bool flag = false;
	int  H, M;
	scanf("%d:%d %c.m", &starthour, &startminute, &chstart);
	scanf("%d:%d %c.m", &endhour,   &endminute,   &chend);
	scanf("%dh%dmin", &hour, &minute);
	if((chstart == 'a' && chend == 'a') || (chstart == 'p' && chend == 'p') || (chstart == 'a' && chend == 'p'))
	{
		TimeProblemAAPPAP(H, M);
	}
	if(chstart == 'p' && chend == 'a')
	{
		TimeProblemPA(H, M);		
	}	
	return 0;	
}

看到别人解后，自愧不如，短短几十行代码，就把所有情况都容纳进去~~

#include <iostream>
using namespace std;

int  h1, h2, m1, m2, hh, mm, h, m;
char a[3], b[3];

int main()
{
	scanf("%d:%d %s\n", &h1, &m1, a);
	scanf("%d:%d %s\n", &h2, &m2, b);
	scanf("%dh%dmin", &hh, &mm);
	if(a[0] != b[0]) h2 += 12;
	h = h2 - h1, m = m2 - m1;
	if(m < 0) m += 60, --h;
	if(h < 0) h += 24;
	if(h == hh && m == mm) puts("YES");
	else puts("NO"), cout << h << "h" << m << "min";
	return 0; 
}

有时是真的得多思考，总结才是。