codeforces Round #258(div2) C解题报告

最新推荐文章于 2022-10-30 20:44:43 发布

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最新推荐文章于 2022-10-30 20:44:43 发布

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原文链接：http://www.cnblogs.com/clnchanpin/p/6916360.html

本文介绍了一道关于足球比赛预测的算法题，比赛中存在三支队伍，错过了一些比赛后，根据朋友提供的胜负差线索，判断是否存在可能让三队获胜次数相同。文章提供了详细的解题思路及代码实现。

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There are n games in a football tournament. Three teams are participating in it. Currently k games had already been played.

You are an avid football fan, but recently you missed the whole k games. Fortunately, you remember a guess of your friend for these kgames. Your friend did not tell exact number of wins of each team, instead he thought that absolute difference between number of wins of first and second team will be d1 and that of between second and third team will be d2.

You don't want any of team win the tournament, that is each team should have the same number of wins after n games. That's why you want to know: does there exist a valid tournament satisfying the friend's guess such that no team will win this tournament?

Note that outcome of a match can not be a draw, it has to be either win or loss.

Input

The first line of the input contains a single integer corresponding to number of test cases t (1 ≤ t ≤ 105).

Each of the next t lines will contain four space-separated integers n, k, d1, d2 (1 ≤ n ≤ 1012; 0 ≤ k ≤ n; 0 ≤ d1, d2 ≤ k) — data for the current test case.

Output

For each test case, output a single line containing either "yes" if it is possible to have no winner of tournament, or "no" otherwise (without quotes).

Sample test(s)

input
5
3 0 0 0
3 3 0 0
6 4 1 0
6 3 3 0
3 3 3 2

output
yes
yes
yes
no
no

Note

Sample 1. There has not been any match up to now (k = 0, d1 = 0, d2 = 0). If there will be three matches (1-2, 2-3, 3-1) and each team wins once, then at the end each team will have 1 win.

Sample 2. You missed all the games (k = 3). As d1 = 0 and d2 = 0, and there is a way to play three games with no winner of tournament (described in the previous sample), the answer is "yes".

Sample 3. You missed 4 matches, and d1 = 1, d2 = 0. These four matches can be: 1-2 (win 2), 1-3 (win 3), 1-2 (win 1), 1-3 (win 1). Currently the first team has 2 wins, the second team has 1 win, the third team has 1 win. Two remaining matches can be: 1-2 (win 2), 1-3 (win 3). In the end all the teams have equal number of wins (2 wins).

题目大意：

有n场比赛，你错过了k场，然后再错过的k长中，一队与二队的胜利差值为d1，二队与三队的胜利差值为d2。如若有可能。三支队伍获胜次数都一样，则输出yes。否则输出no

解法：

比較有趣且复杂的模拟题，我们能够想象成三个柱子，已知第一个柱子跟第二个柱子的高度差的绝对值。第二个柱子跟第三个柱子高度差的绝对值，如今还有n-k个砖（1高度）。要求使得三个柱子高度一样。

首先。给了d1和d2。我们能够枚举一下有那几种基本情况：

1. d1, 0, d2;

2. 0, d1, d1+d2;

3. 0, d2, d1+d2;

4. 0, d1, d1-d2;

5. 0, d2, d2-d1;

当中4和5这两种情况，取决于d1-d2是否为正数，事实上两种是一种情况=_=#。

然后。依据这5种情况，对于每一种情况，x，y。z都必须 >= 0，且 x + y + z <= k && (k - x - y - z) % 3 == 0。这里我们是为了保证我们枚举的情况是否符合基本要求：三个队伍的比赛次数为k。

接下来，我们须要保证的是，(n-k)%3 == 0 && n/3 >= x, y, z)。

代码：

#include <iostream>
#include <cstdio>
#define LL long long

using namespace std;

LL n, k, d1, d2;

bool check(LL x, LL y, LL z) {
	if (x < 0 || y < 0 || z < 0)  return false;
	if (x + y + z > k)  return false;
	if ((k-(x+y+z))%3 != 0)  return false;

	LL tmp = x+y+z+(n-k);
	LL dv = tmp/3;
	LL md = tmp%3;

	if (md == 0 && x <= dv && y <= dv && z <= dv)
		return true;
	else
		return false;
}

void solve() {
	cin >> n >> k >> d1 >> d2;

	if (check(d1, 0, d2))
		printf("yes\n");
	else if (check(0, d1, d1+d2))
		printf("yes\n");
	else if (check(0, d2, d1+d2))
		printf("yes\n");
	else if (check(0, d1, d1-d2))
		printf("yes\n");
	else if (check(0, d2, d2-d1))
		printf("yes\n");
	else
		printf("no\n");
}

int main() {
	int tcase;
	cin >> tcase;

	while (tcase--) {
		solve();
	}
}

转载于:https://www.cnblogs.com/clnchanpin/p/6916360.html