# | Time | Username | Problem | Language | Result | Execution time | Memory |
---|---|---|---|---|---|---|---|
544782 | rainboy | Balloons (CEOI11_bal) | C++98 | 132 ms | 9832 KiB |
This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include <math.h>
#include <stdio.h>
#define N 200000
double min(double a, double b) { return a < b ? a : b; }
int xx[N]; double yy[N], yy_[N];
double f(int i, int j) {
return (xx[j] * yy_[i] - xx[i] * yy_[j]) / (yy_[i] - yy_[j]);
}
int main() {
static int qu[N];
int n, i, cnt;
scanf("%d", &n);
cnt = 0;
for (i = 0; i < n; i++) {
int r, i_;
scanf("%d%d", &xx[i], &r);
while (cnt >= 2 && f(qu[cnt - 2], qu[cnt - 1]) <= xx[i])
cnt--;
i_ = cnt == 0 ? -1 : qu[cnt - 1];
yy[i] = cnt == 0 ? r : min(r, (double) (xx[i] - xx[i_]) * (xx[i] - xx[i_]) / 4 / yy[i_]);
yy_[i] = sqrt(yy[i]);
printf("%f\n", yy[i]);
while (cnt && (yy[qu[cnt - 1]] <= yy[i] || cnt >= 2 && f(qu[cnt - 2], qu[cnt - 1]) <= f(qu[cnt - 1], i)))
cnt--;
qu[cnt++] = i;
}
return 0;
}
Compilation message (stderr)
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