Fox And Snake(Codeforces Round #290 (Div. 2)A)
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题目大意:
给出一个矩阵的行数n和列数m,在这个矩阵内画一条蛇,要求是蛇尾在(1,1),蛇身延伸到(1,m),再下降两行到(3,m),又延伸到(3,1)...
例如:
5 3
###..#####..###
思路分析:
其实就是直接找规律输出。
代码实现:
#include<stdio.h>#include<string.h>char str[55][55];int main(){ int n,m; while(~scanf("%d%d",&n,&m)){ for(int i=0;i<n;i+=2){ for(int j=0;j<m;j++) str[i][j]='#'; } int flag=1; for(int i=1;i<n;i+=2){ if(flag>0){ for(int j=0;j<m-1;j++) str[i][j]='.'; str[i][m-1]='#'; } else{ for(int j=1;j<m;j++) str[i][j]='.'; str[i][0]='#'; } flag=-flag; } for(int i=0;i<n;i++){ for(int j=0;j<m;j++) printf("%c",str[i][j]); printf("\n"); } } return 0;}
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