题解 | #牛牛的水杯#
牛牛的水杯
https://www.nowcoder.com/practice/c196c47c23394bf3bdd4f82a838df6bf
#include<stdio.h> //利用余数来判断是否有多余且不满的一杯
#define PI 3.14
int main() {
int h, r;
scanf("%d %d", &h, &r);
double v = PI * h * r * r * 0.001;
double n = (int)(10.0 / v); //这儿时整型
if(n == (10.0 / v))
printf("%d", (int)n);
else
printf("%d", (int)(n + 1));
return 0;
}#题解#
#define PI 3.14
int main() {
int h, r;
scanf("%d %d", &h, &r);
double v = PI * h * r * r * 0.001;
double n = (int)(10.0 / v); //这儿时整型
if(n == (10.0 / v))
printf("%d", (int)n);
else
printf("%d", (int)(n + 1));
return 0;
}#题解#
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