A. Brain's Photos
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
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.
Examples
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
____________________
题意：一张照片用矩阵n * m 表示，里面有C M Y W G B 几种颜色，判断这张照片是否是黑白照（黑白照只有W G B 三种）
程序应该还是比较好写的，一开始无脑写了一个版本是：
#include<iostream>
using namespace std ;
string s[100];
int main()
{
int n , m ;
cin >> n >> m ;
int flag = 0 ;
for ( int i = 0 ; i < n ; i++ )
{
for ( int j = 0 ; j < m ; j++ )
{
cin>>s[i][j] ;
}
}
for ( int i = 0 ; i < n ; i++ )
{
for ( int j = 0 ; j < m ; j++ )
{
switch(s[i][j])
{
case'C':
case'M':
case'Y':
flag = 1 ; break ;
default : continue ;
}
}
}
if(flag == 1 )
cout << "#Color"<< endl ;
else cout <<"#Black&White" << endl ;
return 0 ;
}
这样的话，大概时间复杂度在1W左右，应该能A，还是runtime error （但是在本机评测的话RE的数据可以在4.6s跑完，不知道是TLE了还是又是代码写的还会导致segment fault） 。
我在这里认为既然本机4.6s能跑完，应该是TLE了。于是怀疑被无数ACMer所诟病的cin cout 的效率，于是用scanf printf改写：
//#include<iostream>
//using namespace std ;
#include<cstdio>
char s[110][110];
int main()
{
int n , m ;
//freopen("2.txt","r",stdin);
scanf("%d %d",&n,&m);
int flag = 0 ;
for ( int i = 0 ; i < n ; i++ )
{
for ( int j = 0 ; j < m ; j++ )
{
//cin>>s[i][j] ;
scanf("%s",&s[i][j]);
switch(s[i][j])
{
case'C':
case'M':
case'Y':
flag = 1 ; break ;
default : continue ;
}
}
}
if(flag == 1 )
//cout << "#Color"<< endl ;
printf("#Color\n");
else //cout <<"#Black&White" << endl ;
printf("#Black&White\n");
return 0 ;
}
成功AC。
但是这个还是有优化空间的，发现开了那么大数组其实根本没必要使用。
最后优化一下代码：
#include<cstdio>
int main()
{
int n , m ;
scanf("%d %d",&n,&m);
int flag = 0 ;
char t[2];
for ( int i = 0 ; i < n ; i++ )
{
for ( int j = 0 ; j < m ; j++ )
{
//cin>>s[i][j] ;
scanf("%s",&t);
switch(t[0])
{
case'C':
case'M':
case'Y':
flag = 1 ; break ;
default : continue ;
}
}
}
if(flag == 1 )
//cout << "#Color"<< endl ;
printf("#Color\n");
else //cout <<"#Black&White" << endl ;
printf("#Black&White\n");
return 0 ;
}
这里似乎还是有字符串的问题，一开始蠢了字符串开成了char t[1]；还错了好几次，貌似我访问t[1]访问到了m的地址，于是t[0]存放字符，t[1]把m的值改变了导致非常奇怪的跑法。
ps：在这里之所以我总是不甘心的用%s 字符串处理，是为了处理空格。
最后 AC记录：