西电复试之——CCF201803-02 碰撞的小球 满分

西电复试之——CCF201803-02 碰撞的小球 满分

添加链接描述

很有意思的一道题
题目见链接

#include<iostream>
using namespace std;

int sudu[101] = { };
int p[101] = {};
int main() {
	int n, L, t;
	cin >> n >> L >> t;
	//规定速度方向向右为1
	for (int i = 0; i < n; i++) {
		cin >> p[i];
		sudu[i] = 1;
		if (p[i] == 0 || p[i] == L) sudu[i] = -sudu[i];
	}
	for (int i = 0; i < t; i++) {
		//在端点换方向
		for (int j = 0; j < n; j++) {
			p[j] = p[j] + sudu[j];
			if (p[j] == 0 || p[j] == L) sudu[j] = -sudu[j];
		}
		for (int j = 0; j < n; j++) {
			for (int k = j + 1; k < n; k++) {
				if (p[k] == p[j]) {
					sudu[j] = -sudu[j];
					sudu[k] = -sudu[k];
				}
			}
			
		}
	}
	for (int i = 0; i < n; i++) {
		cout << p[i] << " ";
	}
	return 0;
}

一开始写错了，是因为审题不清楚，错认为，输入的小球顺序是有序的

借鉴大佬的：解决了我上述问题，加一个结构体数组，记录id，然后排序，思想碰撞很舒服
结构体数组排序解法

#include<iostream>
using namespace std;
#include<algorithm>

const int N = 100;
struct node {
	int id;
	int pos;
	int step;
}b[N];

bool cmp1(node a,node b)
{
	return a.pos < b.pos;
}
bool cmp2(node a, node b) {
	return a.id < b.id;
}

int main() {
    int n, l, t, id = 0;

    cin >> n >> l >> t;
    for (int i = 0; i < n; i++) {
        b[i].id = ++id;
        cin >> b[i].pos;
        // 开始往右走，到达两端则回头
        b[i].step = 1;
        if (b[i].pos == l || b[i].pos == 0)
            b[i].step = -b[i].step;
    }
    sort(b, b + n, cmp1);
    for (int i = 0; i < t; i++) {
        // 走一步
        for (int j = 0; j < n; j++) {
            b[j].pos += b[j].step;
            // 到达两端则回头
            if (b[j].pos == l || b[j].pos == 0)
                b[j].step = -b[j].step;
        }

        // 判断是否碰头，碰头则掉头（排序后只需要比较相邻的）
        for (int j = 1; j < n; j++)
            if (b[j].pos == b[j - 1].pos)
                b[j].step = -b[j].step, b[j - 1].step = -b[j - 1].step;
    }
    sort(b, b + n, cmp2);
    for (int i = 0; i < n; i++)
        cout << b[i].pos << " ";
    return 0;
}