C++ Language used. This assignment is to give you practice using one-dimensional
ID: 3859395 • Letter: C
Question
C++ Language used.
This assignment is to give you practice using one-dimensional arrays and sorting.
In competitive diving, each diver makes dives of varying degrees of difficulty. Nine judges score each dive from 0 through 10 in steps of 0.5. The difficulty is a floating-point value between 1.0 and 3.0 that represents how complex the dive is to perform. The total score is obtained by discarding the lowest and highest of the judges’ scores, adding the remaining scores, and then multiplying that total by the degree of difficulty. Write a program to score each of the dives, using the following input and output specifications, and determine the winner of the competition.
Input:
Create the file m6dive.txt
The first line contains an integer for the number of divers in the competition and subsequent lines contain:
Diver’s name (10 characters max, no blanks included), difficulty (double), and judges’ ratings (nine doubles). There is one line of data for each diver.
Example file:
Output:
The name and difficulty, followed by the scores sorted into increasing order, in tabular form with appropriate headings along with the earned total points for that dive.
Example for sample data above
At the end of the table, print out the name of the winner of the competition (the person with the highest points) and his/her final score.
Hint: Use functions to modularize your program.
Anne 2.0 8.0 8.5 8.5 8.5 9.0 9.0 9.0 9.5 9.5 124.0 Sarah 1.6 7.0 7.5 8.0 8.0 8.0 8.5 8.5 8.5 9.0 91.2Explanation / Answer
#include <stdio.h>
#include <stdlib.h>
#define MAX_DIVER 50
#define MAX_NAMELENGTH 20
#define N_JUDGES 5
int input(FILE* fp, char name[MAX_DIVER][MAX_DIVER], float score[MAX_DIVER][N_JUDGES], float difficulty[MAX_DIVER]);
void sort(float score[MAX_DIVER][N_JUDGES], int diverid);
void calculatescore(float score[MAX_DIVER][N_JUDGES], int diverid, float difficulty[MAX_DIVER], float totalScore[MAX_DIVER]);
void output(char name[MAX_DIVER][MAX_DIVER],float score[MAX_DIVER][N_JUDGES],float totalScore[MAX_DIVER], int diverid);
int main(void)
{
FILE* fp;
char name[MAX_DIVER][MAX_DIVER];
float score[MAX_DIVER][N_JUDGES];
float difficulty[MAX_DIVER];
int divercount;
float totalScore[MAX_DIVER];
int i;
int j;
fp = fopen("DIV1.txt", "r");//open file in read mode
if (fp == NULL)
{
return 1;
}
divercount = input(fp, name, score, difficulty);// this function reads data from input file and calculates the total number of diver
sort(score, divercount);//this function sorts socres from each diver
calculatescore(score, divercount, difficulty, totalScore);// this function calculates total score for each diver
output(name, score, totalScore, divercount);// this function prints output in a file
system("pause");
return 0;
}
/*==========================================input========================================================================================*/
/*
this function reads data from input file and calculates the total number of diver
pre:FILE* fp, char name[MAX_DIVER][MAX_DIVER], float score[MAX_DIVER][N_JUDGES], float difficulty[MAX_DIVER]
post: int diverid
*/
int input(FILE* fp, char name[MAX_DIVER][MAX_DIVER], float score[MAX_DIVER][N_JUDGES], float difficulty[MAX_DIVER])
{
int ch;
int i;
int diverid = 0;
while (1)
{
for (i=0; i<MAX_NAMELENGTH; i++)
{
ch = fgetc(fp);// readging character by character
if(ch == EOF)
{
break;
}
name[diverid][i] = ch;//populating name array
}
if (i != MAX_NAMELENGTH) {
break;
}
fscanf(fp, "%f", difficulty+diverid);// reading population difficulty level array
for (i = 0; i<N_JUDGES; i++)
{
fscanf(fp, "%f", score[diverid]+i);//reading and population score array
}
diverid++;//calculating total number of divers
}
return diverid;
}
/*===========================sort=====================================================*/
/*
this function sorts socres from each diver
pre:float score[MAX_DIVER][N_JUDGES], int diverid
post:
*/
void sort(float score[MAX_DIVER][N_JUDGES], int diverid)
{
int current; int walker;
int smallestIndex;
float temp;
int x;
for(x=0;x<=diverid;x++)//sorting from lowest to highest
{
for (current = 0; current < N_JUDGES - 1; current++)
{
smallestIndex = current;
for (walker = current; walker < N_JUDGES; walker ++)
{
if (score[x][walker] < score[x][smallestIndex])
smallestIndex = walker;
}//for walker
//Swap to position smallest at what is the current position
temp = score[x][current];
score[x][current] = score[x][smallestIndex];
score[x][smallestIndex] = temp;
}//for current
}
return;
}
/*====================================calculatescore============================================*/
/*
this function calculates total score for each diver
pre: float score[MAX_DIVER][N_JUDGES], int diverid, float difficulty[MAX_DIVER],float totalScore[MAX_DIVER]
post:
*/
void calculatescore(float score[MAX_DIVER][N_JUDGES], int diverid, float difficulty[MAX_DIVER],float totalScore[MAX_DIVER])
{
int i;
int j;
float addScore = 0;
for(i=0;i<diverid;i++)
{
for(j=1;j<N_JUDGES-1;j++)//adding scores 2nd lowest to 2nd highest
{
addScore = addScore + score[i][j];
}
totalScore[i] = (addScore/(N_JUDGES-2)) * difficulty[i];//calculation total score
}
return;
}
/*==============================output=================================================*/
/*
this function prints output in a file
pre: char name[MAX_DIVER][MAX_DIVER],float score[MAX_DIVER][N_JUDGES],float totalScore[MAX_DIVER], int diverid
post:
*/
void output(char name[MAX_DIVER][MAX_DIVER],float score[MAX_DIVER][N_JUDGES],float totalScore[MAX_DIVER], int diverid)
{
int i;
int j;
FILE* fp;
fp = fopen("output.txt", "w");//open output file in read mode
fprintf(fp, "name score score score totalscore ");
char ch;
if (fp==NULL)
{
printf("could not open output.txt for writing ");
return;
}
for(j=0;j<diverid;j++)
{
for (i=0; i<MAX_NAMELENGTH; i++)
{
ch = name[j][i];
fputc (ch, fp);
}
fprintf(fp, "%-3.1f %-3.1f %-3.1f %4.2f", score[j][1], score[j][2], score[j][3], totalScore[j]);// primting scores from 2nd lowest to 2nd highest and total score
fprintf(fp, " ");
}
fclose(fp);
}
DIV1.txt
KNIFE JACK 1.3 6.0 5.1 6.3 5.9 6.5
WILLIAMSON FLIP A 1.4 3.5 4.1 4.7 7.2 3.8
SOMMER TODD 1.2 8.0 9.1 8.1 9.3 9.0
SWAN MIKE 1.1 4.3 2.1 9.9 6.2 7.0
sample output
name score score score totalscore
KNIFE JACK 5.9 6.0 6.3 7.89
WILLIAMSON FLIP A 3.8 4.1 4.7 14.37
SOMMER TODD 8.1 9.0 9.1 22.80
SWAN MIKE 4.3 6.2 7.0 27.32
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