提交记录 16826
提交时间 |
评测时间 |
2021-10-25 11:01:45 |
2023-09-03 19:42:03 |
#include <cstdio>
#include <cmath>
using namespace std;
double sqrt(double x);
int main() {
double x = 1000;
printf("%.3lf\n", sqrt(x));
return 0;
}
double sqrt(double x) {
double l = 1, r = x, mid;
for (mid = (l + r) / 2; abs(mid * mid - x) > 1E-5; mid = (l + r) / 2)
if (mid * mid < x)
l = mid;
else
r = mid;
return mid;
}
Compilation | N/A | N/A | Compile OK | Score: N/A | 显示更多 |
Testcase #1 | 7.63 us | 20 KB | Accepted | Score: 100 | 显示更多 |
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