原題鏈接: https://www.nowcoder.com/acm/contest/105/A
空間限制:C/C++ 32768K,其他語言65536K
64bit IO Format: %lld
題目描述
輸入描述:
第一行輸入一個整數n,表示樣例個數。接下來每組樣例兩行,表示開始時間和結束時間,格式爲xdayHH:MM:SS,x是一個整數表示第幾天,0 < x < 20000,HH表示小時,MM表示分鐘,SS表示秒,保證時間合法,結束時間比開始時間晚。
輸出描述:
每組數據輸出一行一個整數,表示經過的秒數。
輸入
2 1day03:26:12 1day04:26:12 123day15:00:01 19999day15:00:00
輸出
3600 1717286399
#include<bits/stdc++.h>
using namespace std;
int bd, bh, bm, bs;
int ed, eh, em, es;
int d, h, m, s;
int t;
int main(){
cin>>t;
while(t--){
scanf("%dday%d:%d:%d", &bd, &bh, &bm, &bs);
scanf("%dday%d:%d:%d", &ed, &eh, &em, &es);
if(bd == ed){
int time1 = bh*3600 + bm*60 + bs;
int time2 = eh*3600 + em*60 + es;
cout<<time2-time1<<endl;
continue;
}
else{
int time1 = bh*3600 + bm*60 + bs;
int time2 = eh*3600 + em*60 + es;
int time3 = ed - bd -1;
time3 = time3*86400;
time1 = 86400 - time1;
cout<<time1+time2+time3<<endl;
}
// cout<<bd<<" "<<bh<<" "<<bm<<" "<<bs<<endl;
}
}
/*
2
1day03:26:12
1day04:26:12
123day15:00:01
19999day15:00:00
*/