UVaOJ 489 - Hangman Judge
来源:互联网 发布:搜狐微博 知乎 编辑:程序博客网 时间:2024/05/06 03:34
#include <stdio.h>#include <string.h>#include <ctype.h>bool isWin(int a[]){ for(int i = 0; i<26; i++) { if(a[i] == 1) return false; } return true;}int main(){ freopen("in.txt","r",stdin); int n; while(scanf("%d",&n)&&n!=-1) { int flag[26]= {0}; memset(flag,0,sizeof(flag)); getchar(); char s[1000],d[1000]; gets(s); int slen = strlen(s); for(int i = 0 ; i<slen; i++) { flag[s[i]-'a'] = 1; } gets(d); int dlen = strlen(d); int error = 0; bool win = false,isLose = false; for(int i = 0; i<dlen; i++) { if(flag[d[i]-'a'] == 1) { flag[d[i]-'a']=2; } else if(flag[d[i]-'a'] == 2||flag[d[i]-'a'] == 3) { continue; } else { flag[d[i]-'a'] = 3; error++; if(error == 7) { isLose = true; break; } } if(isWin(flag)) { win = true; break; } } printf("Round %d\n",n); if(win) { printf("You win.\n"); continue; } if(isLose) { printf("You lose.\n"); continue; } printf("You chickened out.\n"); } return 0;}
这个题花了比较长的 时间,囧~~
1、再提供三组测试数据,下面三种能过,基本就能A了。
4
cheeseab
cheeeeeeeeeeseabing
5
chinese
aaaaaaachinese
6
beautiful
beautiful
2、
上面
第4组数据说的是:同一个字母猜对了多次
第5组数据说的是:同一个字母猜错了多次,只记 错一次
第6组数据说的是:一个很长的数据
0 0
- uvaoj 489 - Hangman Judge
- UVaOJ 489 - Hangman Judge
- UVaOJ 489 - Hangman Judge
- UVaOJ 489 - Hangman Judge
- 489 - Hangman Judge
- 489 - Hangman Judge
- uva-489 - Hangman Judge
- Uva 489: Hangman Judge
- UVa 489 - Hangman Judge
- 489 - Hangman Judge
- UVA 489 - Hangman Judge
- UVA 489 - Hangman Judge
- 489 - Hangman Judge
- 489 - Hangman Judge
- UVA 489 - Hangman Judge
- 489 - Hangman Judge
- uva 489 Hangman Judge
- 489 - Hangman Judge
- 【链表】链表上的绝技
- 嵌入式入门-liunx-shell基础语法
- POJ 3525 Most Distant Point from the Sea 半平面交+二分
- 字符串比较和复制函数
- hdu 2594 java实现字符串KMP算法
- UVaOJ 489 - Hangman Judge
- Flatten Binary Tree to Linked List
- 【美萍超市管理系统】汉码盘点机无缝对接 金蝶盘点机条码数据采集器智能终端PDA
- 非洲小孩
- Number of 1 Bits--LeetCode
- Python对象类型
- 将双向链表转化为二叉堆并有序输出
- zoj 3772 Calculate the Function
- 6. linux文件处理的有用工具