codeforces-16A-A. Flag （C && 简单模拟 && 暴力农夫山泉）

A. Flagtime limit per test2 secondsmemory limit per test64 megabytesinputstandard inputoutputstandard outputAccording to a new ISO standard, a flag of every country should have a chequered field n × m, each square should be of one of10 colours, and the flag should be «striped»: each horizontal row of the flag should contain squares of the same colour, and the colours of adjacent horizontal rows should be different. Berland's government asked you to find out whether their flag meets thenew ISO standard.InputThe first line of the input contains numbers n and m (1 ≤ n, m ≤ 100), n —the amount of rows, m — the amount of columns on the flag of Berland. Then there follows the description of the flag: each of the following n linescontain m characters. Each character is a digit between 0 and 9,and stands for the colour of the corresponding square.OutputOutput YES, if the flag meets the new ISO standard, and NO otherwise.Examplesinput

3 3
000
111
222

output

YES

input

3 3
000
000
111

output

NO

input

3 3
000
111
002

output

NO

题目的意思其实很简单，就是输入,n行m列的这样一个方阵，如果满足，

每行的字符都相等，每列的字符都不相等，则输出YES，否则输出NO.

简单暴力农夫山泉！

代码：

#include<stdio.h>#include<stdlib.h>int main(){char str[105][105];int a,b;scanf("%d%d",&a,&b);getchar();for(int i=0;i<a;i++){for(int j=0;j<b;j++){scanf("%c",&str[i][j]);}//读取回车getchar();}//判断列是否不同int flag1 = 1;//判断行是否相等int flag2 = 1;for(int i=1;i<a;i++){//列相同则falg1为-1if(str[i-1][0]==str[i][0]){flag1 = -1;break;}//行不同则flag2为-1for(int j=1;j<b;j++){if(str[i-1][j-1]!=str[i-1][j]){flag2 = -1;break;}}}//最后一行补上for(int j=1;j<b;j++){if(str[a-1][j-1]!=str[a-1][j]){flag2 = -1;break;}}if(flag1==1&&flag2==1){printf("YES\n");}else{printf("NO\n");}return 0;}