Question

Need this in C

Code is given below

e Dots l lah dit Problem 1. (30 points) Fre bendord.cto obtain the free in decimal representation For ATY. this problem we complete the code in i tive long inte ens (W written the following positive long term 123, 40, 56, 7, 8, 9, 90, 900 the frequencies of all the digits are: 0:4, 1:1, 2:1, 3:1, 4:1, 5:1, 6:1, 7: 1. 8: 1.9: 3 In this example, the free ency of O is because digit appears four time, one in 10. once in 90, and twice in 900. Also, since digit 9 appears once in three numbers, 90, 90, and 900, the frequency for digit 9 is 3. The frequency of other non-ero dicits is Specifically, we need to write following functions in this problem. The first function is void update frequency (unsigned long k, unsigned long freq0)); This function takes an unsigned long and an array freql as input and updates frequ coring to the digits in k. Array freqt has 10 clements, one for each decimal digit freqto store the frequency of o, freq[1] stones the frequency of land so on. If freq hus the values shown in the above example, and the input 9000, the freq should have the following values after the function call. The frequency of digit 9 increws from 3 to 7 The frequency of other digits does not change. 0: 4, 1:1, 2:1, 3:1, 4:1, 5: 1. 6: 1. 7: 1. 8: 1, 9: 7. The second function we need to write is void frequencies(unsigned long all, int n, unsigned long freq[]); This function counts the frequencies of the digits of all n unsigned long integers in array al). The function stores frequencies of digits in array freq... which luas 10 elements, one for each decimal digit. freq[0] stores the frequency of O. freq(1) stores the frequency of 1. and so on. The main() function in the starter code should not be changed. It calls frequencies) twice, on numbers with different distributions. The command line argument specifies the mumber of elements in the array. Below are some sample sessions running the program, $./benford 100 0:11 1:21 2:20 3:20 4:20 5:20 6:20 7 :20 8:20 9:20 0:38 1:57 2:37 3:39 4:30 5:43 6:34 7:22 8:25 9:15 $./benford 1000 0:192 1:301 2:300 3:300 4:300 5:300 6:300 7:300 8:300 9:300 0:573 1:790 2:579 3:592 4:467 5:612 6:556 7:405 8:4409:340 $./benford 100000 0:2893 1:4001 2:4000 3:4000 4:4000 5:4000 6:4000 7:40008:4000 9:4000 0:7769 1:9940 2:7811 3:7877 4:6648 5:8139 6:7531 7:6008 8:64029:5345

Answer 1

#include <stdio.h>
#include<stdlib.h>
#define MAX_N 100000
#define N_DIGITS 10//this value can be anything even we can define through command line
void update_frequency(unsigned long k,unsigned long freq[])//update the frequencies with respect to k
{
int r;
while(k!=0)//take each value
{
r=k%10;
freq[r]++;//increment that value frequency
k=k/10;//next number
}
}
void frequencies(unsigned long a[],int n,unsigned long freq[])
{
for(int i=0;i<n;i++)
{
int t=a[i];//take first value
while(t!=0)//loop through each digit
{
int r=t%10;
freq[r]++;//update that digit frequency
t=t/10;
}
}
}
void print_freq(unsigned long freq[])
{
for(int i=0;i<N_DIGITS;i++)
{
printf("%2d:%-5u",i,freq[i]);//print every frequency value
}
printf("\n");
}
void init_array_uniform(unsigned long a[],int n)
{
for(int i=0;i<n;i++)//this is uniform array it will have numbers in sequence order 1,2,3....
a[i]=i+1;
}
void init_array(unsigned long a[],int n)//it follows some pattern
{
a[0]=1;
for(int i=1;i<n;i++)
a[i]=a[i-1]+i+1;
}
int main(int argc,char **argv)
{
int n=MAX_N;
if(argc>=2)
{
n=atoi(argv[1]);//converting the cmd argument to int
if(n<1||n>MAX_N)
{
printf("The integer must be >=1 and <=%d\n",MAX_N);
return -1;
}
}
unsigned long a[n];
unsigned long freq[N_DIGITS];
for(int i=0;i<N_DIGITS;i++)
freq[i]=(unsigned long)-1>>1;//putting -1 as initial values
init_array_uniform(a,n);//uniform array initialization
frequencies(a,n,freq);//calculate frequencies
print_freq(freq);//print frequencies
printf("updated\n");
update_frequency(6734,freq);//update frequencies
print_freq(freq);//print frequencies
printf("\n");
init_array(a,n);//normal array initialization
frequencies(a,n,freq);//calculate frequencies
print_freq(freq);//print frequencies
printf("updated\n");
update_frequency(43562,freq);//update frequencies
print_freq(freq);//print frequencies

return 0;
}

//I have done with value 10 you can change any value you want in define place or pass from command line arguments

0:38893 1:50000 2:49999 3:49999 4:49999 5:49999 6:49999 7:49999 8:49999 9:49999 updated 0:38893 1:50000 2:49999 3:50000 4:50000 5:49999 6:50000 7:50000 8:49999 9:49999 0:135739 1:141928 2:114020 3:112864 4:106329 5:123121 6:117136 7:106996 8:108963 9:103156 updated 0:135739 1:141928 2:114021 3:112865 4:106330 5:123122 6:117137 7:106996 8:108963 9:103156