IBM | AMAZON | ACCENTURE OA Exams SOLUTIONS
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IPV6 Code
Java
IBM
public static String convertToIpv6(String ipv4Address) {
String[] octets = ipv4Address.split("\\.");
if (octets.length != 4) {
return "Invalid input";
}
if (octets[0].equals("127")) {
return "::1";
}
StringBuilder ipv6 = new StringBuilder("::FFFF:");
try {
for (int i = 0; i < 4; i++) {
int octet = Integer.parseInt(octets[i]);
if (octet < 0 || octet > 255) {
return "Invalid input";
}
String hex = String.format("%02X", octet);
ipv6.append(hex);
if (i == 1) {
ipv6.append(":");
}
}
} catch (NumberFormatException e) {
return "Invalid input";
}
return ipv6.toString();
}
IPV6 Code
Java
IBM
Triangle Code
C++
IBM
double calculate_area(int a, int b, int c) {
double p = (a + b + c) / 2.0;
return (p * (p - a) * (p - b) * (p - c));
}
int compare(const void* a, const void* b) {
triangle* t1 = (triangle*)a;
triangle* t2 = (triangle*)b;
if (t1->area < t2->area)
return -1;
else if (t1->area > t2->area)
return 1;
else
return 0;
}
int sort_by_area(triangle* tr, int n) {
if (tr == NULL || n <= 0) {
return -1;
}
for (int i = 0; i < n; i++) {
tr[i].area = calculate_area(tr[i].a, tr[i].b, tr[i].c);
}
qsort(tr, n, sizeof(triangle), compare);
return 0;
}
Triangle Code
C++
IBM
public static String convertToIpv6(String ipv4Address) {
String[] octets = ipv4Address.split("\\.");
if (octets.length != 4) {
return "Invalid input";
}
if (octets[0].equals("127")) {
return "::1";
}
StringBuilder ipv6 = new StringBuilder("::FFFF:");
try {
for (int i = 0; i < 4; i++) {
int octet = Integer.parseInt(octets[i]);
if (octet < 0 || octet > 255) {
return "Invalid input";
}
String hex = String.format("%02X", octet);
ipv6.append(hex);
if (i == 1) {
ipv6.append(":");
}
}
} catch (NumberFormatException e) {
return "Invalid input";
}
return ipv6.toString();
}
IPV6 Code
Java
IBM
Triangle Code
C++
IBM
https://telegram.me/+Q_kf6B6EFexiNmU9
double calculate_area(int a, int b, int c) {
double p = (a + b + c) / 2.0;
return (p * (p - a) * (p - b) * (p - c));
}
int compare(const void* a, const void* b) {
triangle* t1 = (triangle*)a;
triangle* t2 = (triangle*)b;
if (t1->area < t2->area)
return -1;
else if (t1->area > t2->area)
return 1;
else
return 0;
}
int sort_by_area(triangle* tr, int n) {
if (tr == NULL || n <= 0) {
return -1;
}
for (int i = 0; i < n; i++) {
tr[i].area = calculate_area(tr[i].a, tr[i].b, tr[i].c);
}
qsort(tr, n, sizeof(triangle), compare);
return 0;
}
Triangle Code
C++
IBM
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#include <iostream>
#include <vector>
#include <numeric>
#define MOD 1000000007
using namespace std;
long long mod_inv(long long x, long long mod) {
Β Β Β long long result = 1;
Β Β Β long long power = mod - 2;
Β Β Β while (power) {
Β Β Β Β Β Β Β if (power % 2) {
Β Β Β Β Β Β Β Β Β Β Β result = result * x % mod;
Β Β Β Β Β Β Β }
Β Β Β Β Β Β Β x = x * x % mod;
Β Β Β Β Β Β Β power /= 2;
Β Β Β }
Β Β Β return result;
}
vector<long long> factorial(int n, long long mod) {
Β Β Β vector<long long> fact(n + 1, 1);
Β Β Β for (int i = 2; i <= n; ++i) {
Β Β Β Β Β Β Β fact[i] = fact[i - 1] * i % mod;
Β Β Β }
Β Β Β return fact;
}
long long binomial_coeff(int n, int k, const vector<long long>& fact, long long mod) {
Β Β Β if (k > n || k < 0) {
Β Β Β Β Β Β Β return 0;
Β Β Β }
Β Β Β return fact[n] * mod_inv(fact[k], mod) % mod * mod_inv(fact[n - k], mod) % mod;
}
long long count_ways(int N, int K, const vector<int>& A) {
Β Β Β int sum_A = accumulate(A.begin(), A.end(), 0);
Β Β Β int M = K - sum_A;
Β Β Β vector<long long> fact = factorial(M + N - 1, MOD);
Β Β Β return binomial_coeff(M + N - 1, N - 1, fact, MOD);
}
int main() {
Β Β Β int N, K;
Β Β Β cin >> N >> K;
Β Β Β vector<int> A(N);
Β Β Β for (int i = 0; i < N; ++i) {
Β Β Β Β Β Β Β cin >> A[i];
Β Β Β }
Β Β Β cout << count_ways(N, K, A) << endl;
Β Β Β return 0;
}
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Repost from IBM | AMAZON | ACCENTURE OA Exams SOLUTIONS
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