题目链接

Physics Experiment POJ - 3684

思路

$$ 弹性碰撞问题\\\\ 1.每个球都相当于从高度为h处下落，原因是第i(i 从 0 开始)个球的高度为h+i × 2 × r，\\\\ 下边有i个球，因为下边每有一个球，都相当于下落的距离减少2×r.\\\\ 2.先当r为0求完后再把高度加回去 $$

时间复杂度

$$ O(Nlog(N)) $$

代码

#include <cstdio>
#include <algorithm>
#include <vector>
#include <cmath>

using namespace std;


vector<double> vec;

int g = 10;

double gao(double h, int t) {
    if (t <= 0) {
        return h;
    }
    double t1 = sqrt(2.0 * h / g);
    int k = t / t1;
    if (k & 1) {
        double d = k * t1 + t1 - t;
        return h - g * d * d / 2;
    } else {
        double d = t - k * t1;
        return h - g * d * d / 2;
    }
}

int main() {
    int T;
    scanf("%d", &T);// don't forget &
    while (T--) {
        int n, h, r, t;
        scanf("%d%d%d%d", &n, &h, &r, &t);// don't forget &
        vec.resize(n);
        for (int i = 0; i < n; i++) {
            vec[i] = gao(h, t - i);
        }
        sort(vec.begin(), vec.end());
        for (int i = 0; i < n; i++) {
            printf("%.2f%c", vec[i] + 2.0 * i * r / 100, " \n"[i == n - 1]);
        }
    }

    return 0;
}