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最新推荐文章于 2021-02-05 18:01:51 发布

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本文记录了一次ACM/ICPC世界总决赛的经历，详细描述了比赛中的问题解答过程及提交状态，包括正确解答的问题（A、B、C、G）和错误提交的问题（D）。通过对比赛数据的分析，帮助团队成员更好地理解竞赛表现。

My Summary

Description

 I have a dream that one day in the foyer of the ACM/ICPC World Finals Contest , I could solve all the problems and be the champion . But things always went beyond my mind , we just solved four problems .

 We submitted Problem B at 29 minutes and got “ Accept “ , we passed Problem A at 85 minutes , then we got “Wrong” at 94 minutes on Problem G , but we soon got “Accept” at 98 minutes , then we submitted Problem D but got “Wrong” …..  At last , we passed Problem ABCG , and Problem D was still “Wrong” .

 Now I have to write a summary about the contest , I read our submission states from PC^2 , and I need to list the problem states which illustrates the “Accept” problems and “Wrong” problems , Please help me to do so .

Input

 The first line of input is one integer T ( T<=100 ) , giving the number of test cases in the input . Each test case starts with a line containing a positive integer N ( N<=100 ) , representing the number of the submission states. In the next N lines contains an integer t(0<=t<=300),the submission time.Then a character ( from ‘A’ to ‘J’ ) , the ID of submission problem follows . At last is a string ( “Accept” or “Wrong” ) , the states of this submission .

Output

 For each test case , output two lines , the first line print “Accept” , follow by a colon , a space , then list the name of those accept problems in alphabetical ascending order , if there is no “Accept” problem ( what a shame ! ) , just leaves it blank . The second line is about “Wrong” states , the style of which is similar to the first line . Print a blank line after each test case .

Sample Input

1
10
29 B Accept
85 A Accept
94 G Wrong
98 G Accept
120 D Wrong
170 C Wrong
183 C Accept
190 D Wrong
300 A Accept
300 B Wrong

Sample Output

Accept: ABCG
Wrong: D

 #include<iostream>
using namespace std;
int main()
{
//    freopen("in.txt","r",stdin);
    int n;
    scanf("%d",&n);
    while(n--)
    {
        int times,a,i;
        char s[10],b;
        int ans[10]={0},visit[10]={0};
        scanf("%d",&times);
        for(i=0;i<times;i++)
        {
            scanf("%d %c %s",&a,&b,s);
            if(s[0]=='A' && !visit[b-'A'])
                ans[b-'A']=1,visit[b-'A']=1;
            if(s[0]=='W' && !visit[b-'A'])
                ans[b-'A']=2;
        }
        printf("Accept: ");
        for(i=0;i<10;i++)
        {
            if(ans[i]==1)
                printf("%c",i+'A');
        }
        printf("\n");
        printf("Wrong: ");
        for(i=0;i<10;i++)
        {
            if(ans[i]==2)
                printf("%c",i+'A');
        }
        printf("\n\n");
    }
    return 0;
}