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;

}