UPC5652: Ants on a Circle

时间限制: 2 Sec 内存限制: 256 MB
提交: 54 解决: 15
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题目描述

There is a circle with a circumference of L. Each point on the circumference has a coordinate value, which represents the arc length from a certain reference point clockwise to the point. On this circumference, there are N ants. These ants are numbered 1 through N in order of increasing coordinate, and ant i is at coordinate Xi.

The N ants have just started walking. For each ant i, you are given the initial direction Wi. Ant i is initially walking clockwise if Wi is 1; counterclockwise if Wi is 2. Every ant walks at a constant speed of 1 per second. Sometimes, two ants bump into each other. Each of these two ants will then turn around and start walking in the opposite direction.

For each ant, find its position after T seconds.

Constraints
All input values are integers.
1≤N≤105
1≤L≤109
1≤T≤109
0≤X1<X2<…<XN≤L−1
1≤Wi≤2

输入

The input is given from Standard Input in the following format:

N L T
X1 W1
X2 W2
:
XN WN

输出

Print N lines. The i-th line should contain the coordinate of ant i after T seconds. Here, each coordinate must be between 0 and L−1, inclusive.

样例输入

样例输出

1
3
0

提示

1.5 seconds after the ants start walking, ant 1 and 2 bump into each other at coordinate 1.5. 1 second after that, ant 1 and 3 bump into each other at coordinate 0.5. 0.5 seconds after that, that is, 3 seconds after the ants start walking, ants 1, 2 and 3 are at coordinates 1, 3 and 0, respectively.

来源/分类

AtCoder Grand Contest 013

与经典的蚂蚁问题一样，在两只蚂蚁相遇时视为他们相互穿过，最后n只蚂蚁所处位置的相对顺序不变。

只是这里多了一个新的问题，这是一个圆环，需要锁定一只蚂蚁的位置才可以确定这n个蚂蚁各自的位置。

选择记录第一只蚂蚁的位置，如果有一只蚂蚁顺时针走了一圈，就相当去把第一只蚂蚁往后顶了一个位置，反之就是向前顶了一个位置

#include "bits/stdc++.h"

using namespace std;
typedef long long ll;
const int maxn = 1e5+100;
ll n, l, t;
ll x[maxn];

//如何锁定第一只蚂蚁的位置
int main() {
    freopen("input.txt", "r", stdin);
    cin >> n >> l >> t;
    int w;
    ll temp;
    ll cnt = 0;
    for (int i = 0; i < n; i++) {
        cin >> x[i] >> w;
        if (w == 1) {
            temp = x[i] + t;
            x[i] = temp % l;
            cnt += temp / l;
        } else {
            temp = x[i] - t;
            x[i] = temp % l;
            cnt += temp / l;
            if (x[i] < 0) {
                x[i] += l;
                cnt--;
            }
        }
    }
    sort(x, x + n);
    cnt %= n;
    cnt = (cnt + n) % n;
    for (int i = cnt; i < cnt + n; i++) {
        cout << x[i % n] << endl;
    }
    return 0;

}