从CF1744B学习鸽巢原理等trick

Problem - 1774B - Codeforces

1|0思路

First, We can divide $n$ cells into $\lceil \frac{n}{k}\rceil$ segments that except the last segment, all segments have length $k$. Then in each segment, the colors in it are pairwise different. It's easy to find any $a_i$ should be smaller than or equal to $\lceil \frac{n}{k}\rceil$.

Then we can count the number of $a_i$ which is equal to $\lceil \frac{n}{k}\rceil$. This number must be smaller than or equal to $nmodk$, which is the length of the last segment.

All $a$ that satisfies the conditions above is valid. We can construct a coloring using the method below:

First, we pick out all colors $i$ that $a_i=\lceil \frac{n}{k}\rceil$, then we use color $i$ to color the $j$-th cell in each segment.

Then we pick out all colors $i$ that $a_i<\lceil \frac{n}{k}\rceil -1$ and use these colors to color the rest of cells with cyclic order(i.e. color $j$-th cell of the first segment, of second the segment ... of the $\lceil \frac{n}{k}\rceil$ segment, and let $j+1$. when one color is used up, we begin to use the next color)

At last, we pick out all colors $i$ that $a_i=\lceil \frac{n}{k}\rceil -1$, and color them with the cyclic order.

This method will always give a valid construction.

2|0TRICK

• 长度为 $n$，能分成长度为 $k$​​ 连续段的数量

  • 如 $n=8,k=4$

    • $66666666$

  • $\lceil \frac{n}{k}\rceil$​

    • (n + k - 1) / k//向上取整通法
      

• ((n - 1) % k + 1)//即n % k == 0 ? n : n % k
  

• 鴿巢原理

  • 所以只要有 $a_i>\lceil \frac{n}{k}\rceil$​，就一定会有重复
  • 其他的都放入每个盒子中，如果数量大于 $n$ 即最后一个盒子的数量，还是不行

3|0代码

void solve() {
    int n, m, k;
    std::cin >> n >> m >> k;
    int tot(0);
    int cnt_seg((n + k - 1) / k);
    for (int i = 0, x; i < m; i++) {
        std::cin >> x;
        if (x > cnt_seg) tot = 1 << 30;
        else if (x == cnt_seg) tot++;
    }
    tot <= ((n - 1) % k + 1) ? YES : NO;
}

__EOF__

本文作者：Kdlyh
本文链接：https://www.cnblogs.com/kdlyh/p/18046841.html
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