zoukankan      html  css  js  c++  java
  • HDU1073 Online Judge

    Online Judge

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 5373    Accepted Submission(s): 2054


    Problem Description
    Ignatius is building an Online Judge, now he has worked out all the problems except the Judge System. The system has to read data from correct output file and user's result file, then the system compare the two files. If the two files are absolutly same, then the Judge System return "Accepted", else if the only differences between the two files are spaces(' '), tabs(' '), or enters(' '), the Judge System should return "Presentation Error", else the system will return "Wrong Answer".

    Given the data of correct output file and the data of user's result file, your task is to determine which result the Judge System will return.
     

    Input
    The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.
    Each test case has two parts, the data of correct output file and the data of the user's result file. Both of them are starts with a single line contains a string "START" and end with a single line contains a string "END", these two strings are not the data. In other words, the data is between the two strings. The data will at most 5000 characters.
     

    Output
    For each test cases, you should output the the result Judge System should return.
     

    Sample Input
    4 START 1 + 2 = 3 END START 1+2=3 END START 1 + 2 = 3 END START 1 + 2 = 3 END START 1 + 2 = 3 END START 1 + 2 = 4 END START 1 + 2 = 3 END START 1 + 2 = 3 END
     

    Sample Output
    Presentation Error Presentation Error Wrong Answer Presentation Error
     

    Author
    Ignatius.L
     

    #include <stdio.h>
    #include <string.h>
    
    #define maxn 5010
    
    char str0[maxn], str1[maxn], buf[maxn];
    
    bool isKong(char ch) {
    	return ch == ' ' || ch == '
    ' || ch == '	';
    }
    
    int main() {
    	//freopen("stdin.txt", "r", stdin);
    	int t, mode, id0, id1, i, j;
    	bool PE, WA;
    	scanf("%d", &t);
    	while(t--) {
    		mode = id1 = id0 = 0;
    		PE = WA = false;
    		while(gets(buf)) {
    			if(!strcmp(buf, "START"))
    				continue;
    			else if(!strcmp(buf, "END")) {
    				if(++mode == 2) break;
    				else continue;
    			}
    			if(0 == mode) {
    				for(i = 0; buf[i]; ++i)
    					str0[id0++] = buf[i];
    				str0[id0++] = '
    ';
    				str0[id0] = '';
    			} else {
    				for(i = 0; buf[i]; ++i)
    					str1[id1++] = buf[i];
    				str1[id1++] = '
    ';
    				str1[id1] = '';
    			}
    		}
    		if(id0 != id1) PE = true;
    		for(i = j = 0; i < id0 && j < id1; ) {
    			bool flag = 0;			
    			if(str0[i] != str1[j] && !PE)
    				PE = true;
    			if(isKong(str0[i])) {
    				++i; flag = 1;
    			}
    			if(isKong(str1[j])) {
    				++j; flag = 1;
    			}
    			if(flag) continue;
    			if(str0[i++] != str1[j++]) {
    				if(t == 1) printf("%c%c
    ", str0[i-1]);
    				WA = true; break;
    			}
    		}
    		
    		if(WA) {
    			printf("Wrong Answer
    ");
    			continue;
    		} 
    		while(i < id0) {
    			PE = true;
    			if(isKong(str0[i++]))
    				continue;
    			WA = true; break;
    		}
    		while(j < id1) {
    			PE = true;
    			if(isKong(str1[j++]))
    				continue;
    			WA = true; break;
    		}
    		if(WA) {
    			printf("Wrong Answer
    ");
    			continue;
    		} else if(PE) {
    			printf("Presentation Error
    ");
    			continue;
    		}
    		printf("Accepted
    ");
    	}
    	return 0;
    }


  • 相关阅读:
    ASP.NET Core 基于JWT的认证(二)
    ASP.NET Core 基于JWT的认证(一)
    C#数据Encrypt加密Encrypt解密的算法使用--非对称算法RSACryptoServiceProvider
    C#数据Encrypt加密Encrypt解密的算法使用
    C# 实现winform自动悬浮
    测试winform自动悬浮
    C#实现图像拖拽以及锚点缩放功能
    C# Microsoft.Office.Interop.Owc11 导出excel文件
    Docker修改daemon.json后无法启动的问题
    Kubernetes命名空间
  • 原文地址:https://www.cnblogs.com/yangykaifa/p/6784574.html
Copyright © 2011-2022 走看看