codeforces - 3C - Tic-tac-toe(模拟)
来源:互联网 发布:java的反射如何实现 编辑:程序博客网 时间:2024/05/18 04:27
给一个九宫格棋盘划X和0,判断六种情况的发生情况蛮多的,不过犯了好多纱布的错误,一个if考虑的情况另一个if竟然写掉了!!!写掉了...掉了...了...还有把s[0][2]写成是[0][0]的,我也是醉醉的T T
#include <stdio.h>#include <string.h>#include <math.h>#include <queue>#include <iostream>#include <algorithm>#include <stack>using namespace std;char s[5][5];int heng[5];int shu[5];int main(){ int p1 = 0, p2 = 0, p3 = 0; char s1 = '1', s2 = '2', s3 = '3'; int flag = 0, flag1 = 0, flag2 = 0, flag3 = 0; for(int i = 0; i < 3; i++) scanf("%s", s[i]); if(s[0][0] == s[1][1] && s[0][0] == s[2][2] && s[0][0] != '.') { flag3++; if(s3 != 'X') s3 = s[0][0]; else s3 = 'X'; } if(s[0][2] == s[1][1] && s[0][2] == s[2][0] && s[0][2] != '.') { flag3++; if(s3 != 'X') s3 = s[0][2]; else s3 = 'X'; } memset(heng, 0, sizeof(heng)); memset(shu, 0, sizeof(shu)); for(int i = 0; i < 3; i++) { for(int j = 0; j < 3; j++) { if(s[i][j]!='X' && s[i][j]!='0' && s[i][j]!='.') { flag = 1; break; } if(s[i][j] == 'X') p1++; if(s[i][j] == '0') p2++; if(s[i][j] == '.') p3++; if(i > 0) { if(s[i][j] == s[i-1][j] && s[i][j] != '.') { shu[j]++; if(shu[j] == 2) { flag2++; if(s2 != 'X') s2 = s[i][j]; else s2 = 'X'; } } } if(j > 0) { if(s[i][j] == s[i][j-1] && s[i][j] != '.') { heng[i]++; if(heng[i] == 2) { flag1++; if(s1 != 'X') s1 = s[i][j]; else s1 = 'X'; } } } } } //printf("flag1=%d, flag2=%d, flag3=%d, p1=%d, p2=%d, p3=%d\n",flag1,flag2,flag3,p1,p2,p3); // cout<<s1<<" "<<s2<<" "<<s3<<endl; if(flag == 1) printf("illegal\n"); else if((p1 > p2+1) || (p2 > p1)) { printf("illegal\n"); } else if(p3==9 || (p3==0 && flag1==0 && flag2==0 && flag3==0)) printf("draw\n"); else if(/*(flag1!=0&&flag2!=0) || (flag1!=0&&flag3!=0) || (flag2!=0&&flag3!=0)) ||*/ flag1>1 || flag2>1) printf("illegal\n"); else if(flag1) { if(s1=='X' && p1==p2+1) printf("the first player won\n"); else if(s1=='0' && p1==p2) printf("the second player won\n"); else printf("illegal\n"); } else if(flag2) { if(s2=='X' && p1==p2+1) printf("the first player won\n"); else if(s2=='0' && p1==p2) printf("the second player won\n"); else printf("illegal\n"); } else if(flag3) { if(s3=='X' && p1==p2+1) printf("the first player won\n"); else if(s3=='0' && p1==p2) printf("the second player won\n"); else printf("illegal\n"); } else if(p1 == p2+1) printf("second\n"); else if(p2 == p1) printf("first\n"); return 0;}
0 0
- codeforces - 3C - Tic-tac-toe(模拟)
- CodeForces 3C Tic-tac-toe(模拟)
- cf 3C Tic-tac-toe(模拟)
- codeforces 3C. Tic-tac-toe
- codeforces 3C Tic-tac-toe
- Codeforces 3C. Tic-tac-toe
- Codeforces 3 C. Tic-tac-toe
- CodeForces 3C-Tic-tac-toe
- Codeforces 3C Tic-tac-toe
- C. Tic-tac-toe【模拟】
- [Codeforces]C. Tic-tac-toe
- Codeforces Beta Round #3 / 3C Tic-tac-toe (超级模拟)
- Codeforces Beta Round #3C. Tic-tac-toe
- Codeforces Beta Round #3 C. Tic-tac-toe
- codeforces 3C Tic-tac-toe (想法题)
- CodeForces 3C Tic-tac-toe 井字棋盘游戏
- CodeForces 3C---Tic-tac-toe--思维题
- Codeforces Beta Round #3-C. Tic-tac-toe
- UI笔记:UILabel、UIButton和UITextField
- iOS 8 毛玻璃效果(模糊)
- 泛型(java基础)
- 经典排序算法1--冒泡排序
- Spring Security搭配hibernate,Mysql
- codeforces - 3C - Tic-tac-toe(模拟)
- Flickr Architecture
- 矩阵相乘这个代码也应该是需要掌握的内容,今天一大早就写了一下
- Java技术1-线程池
- 人格的五因素模型
- 桌面虚拟化实施遇到的基本问题
- 细数Python与C++的区别(更新中……)
- 老外被中国app惊呆了
- Diskpart工具应用两则:MBR/GPT分区转换 & 基本/动态磁盘转换