HDU 4178 模拟

Roll-call in Woop Woop High

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 188    Accepted Submission(s): 133

Problem Description

The
new principal of Woop Woop High is not satisfied with her pupils
performance.She introduced a new roll-call process to get a daily
measure of the pupils' learning, which proceeds as follows:At the
beginning of the daily roll-call period each pupil is handed a question,
which they must attempt to answer, before proceeding to their classes. A
pupil stops after the question is answered correctly. Each pupil is
allowed up to five attempts to answer the question correctly. Pupils who
answer correctly on the first attempt are marked present. Pupils who
answer correctly after more than one attempt are encouraged to work at
home. Pupils who fail to develop a correct answer within five attempts
are given remedial classes after school. Pupils who do not give any
answer are marked as abscent. Your task is to write a program that reads
the pupils' assessments and generates performance reports for the
principal to proceed with appropriate actions.

 

Input

The
input starts with an integer K (1 <= K <= 100) indicating the
number of classes. Each class starts with an integer N (1 <= N <=
50) indicating the number of pupils in the class. Each of the following N
lines starts with a pupil's name followed by up-to five assessments of
his/her answers. An assessment of 'yes' or 'y' indicates a correct
answer and an assessment of 'n' or 'no' indicates a wrong answer. A
pupil's name consists of a single string with no white spaces.

 

Output

The attendance report for each class consists of five lines.
The first line consists of the sentence: "Roll-call: X", where X indicates the class number starting with the value of one.
The
second line consists of the sentence: ''Present: Y1 out of N'', where
Y1 is the number of pupils who did not submit a wrong answer.
The
third line consists of the sentence: ''Needs to study at home: Y2 out of
N'', where Y2 is the number of pupils who submitted a number of wrong
answers before submitting the correct answer.
The fourth line
consists of the sentence: ''Needs remedial work after school: Y3 out of
N'', where Y3 indicates the number of pupils whose submitted five wrong
answers.
The fifth line consists of the sentence: ''Absent: Y4 out of N'', where Y4 indicates the number of absent pupils.

 

Sample Input

2
5
Doc n y
sneezy n n no yes
princecharming no n no no n
goofy yes
grumpy n y
5
evilemperor n y
princesslia
r2d2 no no y
obeyonecanopy n no y
darthvedar y

 

Sample Output

Roll-call: 1
Present: 1 out of 5
Needs to study at home: 3 out of 5
Needs remedial work after school: 1 out of 5
Absent: 0 out of 5
Roll-call: 2
Present: 1 out of 5
Needs to study at home: 3 out of 5
Needs remedial work after school: 0 out of 5
Absent: 1 out of 5

 

Source

The 2011 South Pacific Programming Contest

 

//输出的一共5行，第一行略微有点坑，第一行其实就是指的第几组数据，大概就是case的意思

Recommend

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
using namespace std;

struct pp{
    char m[];
    int flag1;
    int cro,wro,sum;
}a[];

int main()
{
    int t;
    scanf("%d",&t);
    int ans1 = ;
    while(t--)
    {
        int n;
        scanf("%d",&n);
        getchar();
        int max1 = ;
        for(int i = ;i < n;i++)
        {
            gets(a[i].m);
            int len = strlen(a[i].m);
            int flag = -;
            int j = ;
            a[i].cro = ;
            a[i].wro = ;
            a[i].sum = ;
            for( ;j < len;j++)
            {
                if(a[i].m[j] == ' ')
                    break;
            }
            j++;
            for( ;j < len;j++)
            {
                if(a[i].m[j] == 'y')
                {
                    flag = ;
                    a[i].cro++;
                    if(a[i].m[j+] == 'e')
                        j = j+;
                    else
                        j++;
                    break;
                }
                else if(a[i].m[j] == 'n')
                {
                    flag = ;
                    a[i].wro++;
                    if(a[i].m[j+] == 'o')
                        j = j+;
                    else
                        j++;
                    break;
                }
            }

            a[i].flag1 = flag;
            j++;
            for( ;j < len;j++)
            {
                if(a[i].m[j] == 'y')
                {
                    a[i].cro++;
                    if(a[i].m[j+] == 'e')
                        j = j+;
                    else
                        j++;
                }
                else if(a[i].m[j] == 'n')
                {
                    a[i].wro++;
                    if(a[i].m[j+] == 'o')
                        j = j+;
                    else
                        j++;
                }
            }
            a[i].sum = a[i].cro + a[i].wro;
            if(max1 < a[i].sum)
                max1 = a[i].sum;
        }

       // for(int i = 0;i < n;i++)
       //     printf("---%s %d %d %d\n",a[i].m,a[i].cro,a[i].wro,a[i].sum);

        int ans2 = ,ans3 = ,ans4 = ,ans5 = ;
        for(int i = ;i < n;i++)
        {
            if(a[i].wro ==  && a[i].cro > )
                ans2++;
            if(a[i].flag1 ==  && a[i].cro > )
                ans3++;
            if(a[i].wro >= )
                ans4++;
            if(a[i].sum == )
                ans5++;
        }
        printf("Roll-call: %d\n",++ans1);
        printf("Present: %d out of %d\n",ans2,n);
        printf("Needs to study at home: %d out of %d\n",ans3,n);
        printf("Needs remedial work after school: %d out of %d\n",ans4,n);
        printf("Absent: %d out of %d\n",ans5,n);
    }
    return ;
}