题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6556
题意：就是一个时区转换和上下午判别。
解题心得：就是一个常规题，时间转换之后时钟大于等于24代表是在Tomorrow，如果小于0代表在Yesterday。只不过有一点就是半夜0点写成12点AM，这里要注意输入和输出，还有就是输出小时是%d，分钟是d。
#include <bits/stdc++.h>
using namespace std
;
typedef long long ll
;
map
<string
, int> maps
;
void init() {
string st
;
st
= "Beijing";
maps
[st
] = 8;
st
= "Washington";
maps
[st
] = -5;
st
= "London";
maps
[st
] = 0;
st
= "Moscow";
maps
[st
] = 3;
}
int main() {
init();
int t
; scanf("%d", &t
);
int T
= t
;
while(t
--) {
int h
, m
;
string now
, Next
, w
;
scanf("%d:%d", &h
, &m
);
cin
>>w
>>now
>>Next
;
if(w
== "PM") {
if(h
!= 12)
h
+= 12;
} else {
if(h
== 12)
h
-= 12;
}
int pos1
, pos2
;
pos1
= maps
[now
];
pos2
= maps
[Next
];
if(pos2
<= pos1
) {
h
-= (pos1
- pos2
);
} else {
h
+= (pos2
- pos1
);
}
printf("Case %d: ", T
-t
);
if(h
< 0) {
printf("Yesterday ");
h
+= 24;
} else if(h
>= 24) {
printf("Tomorrow ");
h
-= 24;
} else {
printf("Today ");
}
string st
;
if(h
< 12) {
st
= "AM";
if(h
== 0) h
+= 12;
}
else {
st
= "PM";
if(h
> 12)
h
-= 12;
}
printf("%d:d %s\n", h
, m
, st
.c_str());
}
return 0;
}
