【CodeForces 707A】Brain's Photos(水)

小鼠Brain在被夏季反派学校拒绝后决定成为一名摄影师。他拍摄了许多彩色和黑白照片,但混合存放导致整理困难。现在需要判断一张图片是否为彩色照片。通过扫描图片矩阵中的每个像素来确定是否存在彩色元素。

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Brain's Photos

Small, but very brave, mouse Brain was not accepted to summer school of young villains. He was upset and decided to postpone his plans of taking over the world, but to become a photographer instead.

As you may know, the coolest photos are on the film (because you can specify the hashtag #film for such).

Brain took a lot of colourful pictures on colored and black-and-white film. Then he developed and translated it into a digital form. But now, color and black-and-white photos are in one folder, and to sort them, one needs to spend more than one hour!

As soon as Brain is a photographer not programmer now, he asks you to help him determine for a single photo whether it is colored or black-and-white.

Photo can be represented as a matrix sized n × m, and each element of the matrix stores a symbol indicating corresponding pixel color. There are only 6 colors:

  • 'C' (cyan)
  • 'M' (magenta)
  • 'Y' (yellow)
  • 'W' (white)
  • 'G' (grey)
  • 'B' (black)

The photo is considered black-and-white if it has only white, black and grey pixels in it. If there are any of cyan, magenta or yellow pixels in the photo then it is considered colored.

Input

The first line of the input contains two integers n and m (1 ≤ n, m ≤ 100) — the number of photo pixel matrix rows and columns respectively.

Then n lines describing matrix rows follow. Each of them contains m space-separated characters describing colors of pixels in a row. Each character in the line is one of the 'C', 'M', 'Y', 'W', 'G' or 'B'.

Output

Print the "#Black&White" (without quotes), if the photo is black-and-white and "#Color" (without quotes), if it is colored, in the only line.

Example
Input
2 2
C M
Y Y
Output
#Color
Input
3 2
W W
W W
B B
Output
#Black&White
Input
1 1
W
Output
#Black&White

思路:有C,M,Y输出color
#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
    int n,m;
    char c;
    bool f=false;
    scanf("%d%d",&n,&m);
    for (int i=0; i<n; i++)
        for (int j=0; j<m; j++){
            getchar();
            scanf("%c",&c);
            if(c=='C' || c=='M' || c=='Y') f=true;
        }
    if(f) printf("#Color\n");
    else printf("#Black&White\n");
    return 0;
}


### 关于 Codeforces 上 'Neo's Escape' 的解决方案 在解决 Codeforces 平台上的 `'Neo's Escape'` 问题时,通常需要考虑图论中的最短路径算法以及动态规划的应用。以下是对此类问题可能涉及的核心概念和技术的分析: #### 图论与最短路径 该问题可以被建模为在一个加权有向图中寻找从起点到终点的最优路径。Dijkstra 算法是一种常用的方法来计算单源最短路径[^1]。如果边权重均为正数,则 Dijkstra 是一种高效的选择。 ```python import heapq def dijkstra(graph, start): distances = {node: float('inf') for node in graph} distances[start] = 0 priority_queue = [(0, start)] while priority_queue: current_distance, current_node = heapq.heappop(priority_queue) if current_distance > distances[current_node]: continue for neighbor, weight in graph[current_node].items(): distance = current_distance + weight if distance < distances[neighbor]: distances[neighbor] = distance heapq.heappush(priority_queue, (distance, neighbor)) return distances ``` #### 动态规划优化 对于某些变体问题,除了简单的最短路径外,还需要引入状态转移方程来进行进一步优化。例如,在存在多种属性约束的情况下(如时间、能量),可以通过定义多维数组 `dp[i][j]` 来表示到达节点 i 使用 j 单位资源所需的最小代价[^2]。 #### 讨论与实现细节 社区内的讨论往往围绕如何处理特殊边界条件展开,比如是否存在负环路或者超大数据集下的性能调优等问题。此外,部分参赛者会分享他们关于数据结构选取的经验教训,例如优先队列 vs. 堆栈的不同适用场景[^3]。
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