题目:给你一些题目的输出结果,推断是AC,PE还是WA。

分析:模拟。

依照题意模拟就可以,注意PE条件为全部数字字符出现顺序同样就可以。

说明:想起非常多年前写的OJ的后台判题程序了╮(╯▽╰)╭。

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <cmath>

using namespace std;

char list1[101][122];
char list2[101][122];
char numb1[12001];
char numb2[12001];

int main()
{
	int n,m,t = 1;
	while (~scanf("%d",&n) && n) {
		getchar();
		for (int i = 0; i < n; ++ i)
			gets(list1[i]);
		scanf("%d",&m);
		getchar();
		for (int i = 0; i < m; ++ i)
			gets(list2[i]);
		
		int AC = 1;
		if (m == n) {
			for (int i = 0; i < n; ++ i)
				if (strcmp(list1[i], list2[i])) {
					AC = 0;
					break;
				}
		}else AC = 0;
		
		int PE = 1,save1 = 0,save2 = 0;
		for (int i = 0; i < n; ++ i)
			for (int j = 0; list1[i][j]; ++ j)
				if (list1[i][j] >= '0' && list1[i][j] <= '9')
					numb1[save1 ++] = list1[i][j];
		for (int i = 0; i < m; ++ i) 
			for (int j = 0; list2[i][j]; ++ j)
				if (list2[i][j] >= '0' && list2[i][j] <= '9')
					numb2[save2 ++] = list2[i][j];		
		if (save1 == save2) {
			for (int i = 0; i < save1; ++ i)
				if (numb1[i] != numb2[i]) {
					PE = 0;
					break;
				}
		}else PE = 0;
		
		printf("Run #%d: ",t ++);
		if (AC) printf("Accepted\n");
		else if (PE) printf("Presentation Error\n");
		else printf("Wrong Answer\n");
	}
    return 0;
}