Extreme Fizz Buzz Fizz Buzz is a classic Computer Science problem about Selection. For an integer x, print: • “Fizz” if x is only divisible by 3, • “Buzz” if x is only divisible by 5, • “FizzBuzz” if x is divisible by both 3 and 5, • the number x if it is not divisible by both 3 and 5. This problem is the extreme version of Fizz Buzz. You are given a list of K distinct integers A and a list of K characters S. Both lists are numbered from 1 to K. The integer Ai corresponds to the character Si. Denote B as the subset of A. There will be a rule for each possible subset: if an integer x is only divisible by all elements in B, print a string which is the concatenation of the corresponding character of each element in B, sorted by the index in ascending order. However, if B is an empty subset, print the first digit of x instead. Does it look complicated? Do not worry, as it is just a generalization of the regular Fizz Buzz. For example, if A = [3, 5] and S = [F, B], you will get the normal Fizz Buzz problem. You can also check the Sample Test Cases for further understanding. Your task is simple. Given the array A and S, with all rules from the subset of A, printthe Extreme Fizz Buzz from 1 to N. Format Input The first line consists of 2 integers N and K. The second line consists of a string S which has exactly K lower-case latin alphabet characters. The third line consists of K integers A1, A2, . . . , AK. Format Output Output N lines. The i-th line is the Extreme Fizz Buzz of i, where i is integer from 1 to N. Constraints • 1 ≤ N, K ≤ 50000 • 1 ≤ Ai ≤ 50000 • It is guaranteed that the integers in A are distinct • It is guaranteed that the characters in S are upper-case latin alphabet characters Sample Input 1 (standard input) 15 2 FB 3 5 Sample Output 1 (standard output) 1 2 F 4 B F 7 8 F B 1 F 3 4 FB Sample Input 2 (standard input) 15 2 BF 3 5 Sample Output 2 (standard output) 1 2 B 4 F B 7 8 B F 1 B 3 4 BF Sample Input 3 (standard input) 13 3 JLB 2 4 3 Sample Output 3 (standard output) 1 J B JL 5 JB 7 JL B J 1 JLB 3 Explanation In the first sample, the given rule is similar to the regular Fizz Buzz. However, print “F” instead of “Fizz”, print “B” instead of “Buzz”, print “FB” instead of “FizzBuzz”, and print the first digit of x instead of the number x. In the second sample, the given rule is similar to the regular Fizz Buzz. However, print “B” instead of “Fizz”, print “F” instead of “Buzz”, and print “BF” instead of “FizzBuzz”, print the first digit of x instead of the number x. In the third sample, 2 corresponds to “J”, 4 corresponds to “L”, and 3 corresponds to “B”. Using the given A and S, there are exactly 8 rules, which are also the number of subset in A: • If x is only divisible by 2, print “J” • If x is only divisible by 4, print “L” • If x is only divisible by 3, print “B” • If x is only divisible by 2 and 4, print “JL” • If x is only divisible by 2 and 3, print “JB” • If x is only divisible by 4 and 3, print “LB” • If x is only divisible by 2, 4, and 3, print “JLB” • If x is not divisible by 2, 4, and 3, print the first digit of x instead #include int main(){ long long int n, k; scanf("%lld %lld\n", &n, &k); char S[55000];long long int A[55000];char ch; for (long long int a = 0; a < k; a++){ scanf("%c", &S[a]); } for (long long int a = 0; a < k; a++){ scanf("%lld", &A[a]); } for (long long int a = 1; a <= n; a++){ char subset[k + 1]; long long int index = -1; for (long long int b = 0; b < k; b++){ if ((a % A[b]) == 0) { index++; subset[index] = S[b]; } } if (index == -1){ printf("%lld\n", a % 10); } else{ for(long long int b = 0; b <= index; b++){ printf("%c", subset[b]); } printf("\n"); } } return 0; } Sir please fix this code by making the looping is only 4 looping, not
Extreme Fizz Buzz
Fizz Buzz is a classic Computer Science problem about Selection. For an integer x, print:
• “Fizz” if x is only divisible by 3,
• “Buzz” if x is only divisible by 5,
• “FizzBuzz” if x is divisible by both 3 and 5,
• the number x if it is not divisible by both 3 and 5.
This problem is the extreme version of Fizz Buzz. You are given a list of K distinct integers A and a list of K characters S. Both lists are numbered from 1 to K. The integer
x is only divisible by all elements in B, print a string which is the concatenation of the corresponding character of each element in B, sorted by the index in ascending order. However, if B is an empty subset, print the first digit of x instead. Does it look complicated? Do not worry, as it is just a generalization of the regular Fizz Buzz. For example, if A = [3, 5] and S = [F, B], you will get the normal Fizz Buzz problem. You can also check the Sample Test Cases for further understanding. Your task is simple. Given the array A and S, with all rules from the subset of A, printthe Extreme Fizz Buzz from 1 to N.
Format Input
The first line consists of 2 integers N and K. The second line consists of a string S which has exactly K lower-case latin alphabet characters. The third line consists of K integers A1, A2, . . . , AK.
Format Output
Output N lines. The i-th line is the Extreme Fizz Buzz of i, where i is integer from 1 to N.
Constraints
• 1 ≤ N, K ≤ 50000
• 1 ≤ Ai ≤ 50000
• It is guaranteed that the integers in A are distinct
• It is guaranteed that the characters in S are upper-case latin alphabet characters
Sample Input 1 (standard input)
15 2
FB
3 5
Sample Output 1 (standard output)
1
2
F
4
B
F
7
8
F
B
1
F
3
4
FB
Sample Input 2 (standard input)
15 2
BF
3 5
Sample Output 2 (standard output)
1
2
B
4
F
B
7
8
B
F
1
B
3
4
BF
Sample Input 3 (standard input)
13 3
JLB
2 4 3
Sample Output 3 (standard output)
1
J
B
JL
5
JB
7
JL
B
J
1
JLB
3
Explanation
In the first sample, the given rule is similar to the regular Fizz Buzz. However, print “F” instead of “Fizz”, print “B” instead of “Buzz”, print “FB” instead of “FizzBuzz”, and print the first digit of x instead of the number x.
In the second sample, the given rule is similar to the regular Fizz Buzz. However, print “B” instead of “Fizz”, print “F” instead of “Buzz”, and print “BF” instead of “FizzBuzz”, print the first digit of x instead of the number x.
In the third sample, 2 corresponds to “J”, 4 corresponds to “L”, and 3 corresponds to “B”. Using the given A and S, there are exactly 8 rules, which are also the number of subset in A:
• If x is only divisible by 2, print “J”
• If x is only divisible by 4, print “L”
• If x is only divisible by 3, print “B”
• If x is only divisible by 2 and 4, print “JL”
• If x is only divisible by 2 and 3, print “JB”
• If x is only divisible by 4 and 3, print “LB”
• If x is only divisible by 2, 4, and 3, print “JLB”
• If x is not divisible by 2, 4, and 3, print the first digit of x instead
#include <stdio.h>
int main(){
long long int n, k;
scanf("%lld %lld\n", &n, &k);
char S[55000];long long int A[55000];char ch;
for (long long int a = 0; a < k; a++){
scanf("%c", &S[a]);
}
for (long long int a = 0; a < k; a++){
scanf("%lld", &A[a]);
}
for (long long int a = 1; a <= n; a++){
char subset[k + 1];
long long int index = -1;
for (long long int b = 0; b < k; b++){
if ((a % A[b]) == 0) {
index++;
subset[index] = S[b];
}
}
if (index == -1){
printf("%lld\n", a % 10);
}
else{
for(long long int b = 0; b <= index; b++){
printf("%c", subset[b]);
}
printf("\n");
}
}
return 0;
}
Sir please fix this code by making the looping is only 4 looping, not 5 looping, thx.
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