思路:
先处理后两位,直接加30然后逢60进一,比10小就输出0
再处理前两位,直接加3然后逢8进一,比10小就输出0
C o d e Code Code:
#include <cstdio>
#include <iostream>
using namespace std;
char str;
int sum1,sum2,ans1,ans2;
int main ()
{
cin>>str;
sum1 = sum1 * 10 + str - '0';//转换为数字
cin>>str;
sum1 = sum1 * 10 + str - '0';
cin>>str;
cin>>str;
sum2 = sum2 * 10 + str - '0';
cin>>str;
sum2 = sum2 * 10 + str - '0';
ans2 = sum2 + 30;//后两位
ans1 = sum1 + ans2 / 60;
ans2 %= 60;
ans1 += 3;
ans1 %= 24;
if(ans1 < 10)printf("0");
printf("%d:",ans1);
if(ans2 < 10)printf("0");//输出0
printf("%d",ans2);
return 0;
}