jilok @ 2024-08-26 06:47:31
int main() {
```c
int main() { int s, v; scanf("%d %d", &s, &v); if (s < 1 && v>10000 && v >= 0) {
}
else
{
int i = 480 - 10 - s / v;
if (i == 0) {
printf("00:00");
}
else if (i > 0) {
int q = s / v;
int w = s % v;
if (w > 0) {
q++;
}
int e = 480 - q - 10;
int h = e / 60;
int m = e % 60;
if (h < 10) {
printf("0%d:%d",h,m);
}
else
{
printf("%d:%d", h, m);
}
}
else if (i < 0 && i >= -960) {
int r = s / v;
int t = s % v;
if (t > 0) {
r++;
}
int y = r - 480 + 10;
int H = 23 - y / 60;
int M = 60 - y % 60;
if (H < 10) {
printf("0%d:%d", H, M);
}
else
{
printf("%d:%d", H, M);
}
}
else
{
}
}
return 0;
}
int s, v;
scanf("%d %d", &s, &v);
if (s < 1 && v>10000 && v >= 0) {
}
else
{
int i = 480 - 10 - s / v;
if (i == 0) {
printf("00:00");
}
else if (i > 0) {
int q = s / v;
int w = s % v;
if (w > 0) {
q++;
}
int e = 480 - q - 10;
int h = e / 60;
int m = e % 60;
if (h < 10) {
printf("0%d:%d",h,m);
}
else
{
printf("%d:%d", h, m);
}
}
else if (i < 0 && i >= -960) {
int r = s / v;
int t = s % v;
if (t > 0) {
r++;
}
int y = r - 480 + 10;
int H = 23 - y / 60;
int M = 60 - y % 60;
if (H < 10) {
printf("0%d:%d", H, M);
}
else
{
printf("%d:%d", H, M);
}
}
else
{
}
}
return 0;
}
by Jason_Teng @ 2024-08-26 08:19:46
这题忒简单了!,你当7:50到校,再写:
int u = s / v + (s / v * v < v ? 1 : 0) + 10;
int h = u / 60 + (u % 60 != 0 ? 1 : 0);
int m = 60 - u % 60;
if(m == 60)
m = 0;
printf("0%d:%d",h,m);
就行啦
by Jason_Teng @ 2024-08-26 08:20:55
还有,新手蒟蒻应用标配cout
by Jason_Teng @ 2024-08-26 08:21:35
@a3481171079
by Jason_Teng @ 2024-08-26 08:22:21
其实我也用cout
by Jason_Teng @ 2024-08-26 08:30:43
条件表达式详解
by jilok @ 2024-08-26 17:29:09
@shabiJason 大佬,我还没学cout
by jilok @ 2024-08-26 17:29:29
@shabiJason 感谢感谢
by Jason_Teng @ 2024-08-26 20:23:16
壶关? @a3481171079
by Jason_Teng @ 2024-08-26 20:24:51
@a3481171079 ,cout详解明天好
by jilok @ 2024-08-28 01:19:11
@shabiJason 好的