B1026

    技术2024-10-14  52

    程序运行时间 (15分)

    总结:

    四舍五入判定:①round函数 (直接,但涉及数据类型转化,麻烦)

                              ②根据题意,(c2-c1)/100四舍五入即余数满50进一,不满50舍去,令time= ((c2-c1) + 50) / 100即可。

    代码:

    #include<cstdio> #include<cmath> const int CLK_TCK = 100; int main() { int c1, c2; scanf("%d%d", &c1,&c2); int ticks = c2-c1; double time = (double)ticks/CLK_TCK; int hour,min,sec; hour = time/3600; min = (time-3600*hour)/60; sec = round(time-3600*hour-60*min);//四舍五入 //或者直接算 int time= ((c2-c1) + 50) / 100; //或者if判断 (c2-c1)%100 >= 50 printf("%02d:%02d:%02d", hour,min,sec); return 0; }

     

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