๐—–๐—ฆ ๐—”๐—น๐—ด๐—ผ ๐Ÿ’ป ๐ŸŒ ใ€Ž๐—–๐—ผ๐—บ๐—ฝ๐—ฒ๐˜๐—ถ๐˜๐—ถ๐˜ƒ๐—ฒ ๐—ฃ๐—ฟ๐—ผ๐—ด๐—ฟ๐—ฎ๐—บ๐—บ๐—ถ๐—ป๐—ดใ€
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Examity hiring Process Associate Trainee ๐Ÿ’ช๐Ÿ’ช๐Ÿฅณ๐Ÿฅณ

Details
Job Role: Process Associate Trainee
Qualifications: Any Graduate
Experience: 0 โ€“ 2 years
Job Type: Full Time
Location: Hyderabad
Salary: 5 LPA

Skills/Requirements:
1. Ability to accurately follow instructions and meticulously handle tasks, ensuring data accuracy and process compliance.
2. Eagerness to absorb new information, learn industry-specific processes, and quickly adapt to the roleโ€™s demands.
3. Proficiency in using office software such as Microsoft Office Suite (Excel, Word) and familiarity with data entry and management tools.
4. Capability to manage time effectively, prioritize tasks, and meet deadlines in a fast-paced work environment.

For Hyderabad - Come directly to the Venue

Date of Interview: 17th August

Time: 10.00 AM โ€“ 4.30 PM

Address: Examity India ON-LINE Proctoring LLP. 2nd Floor , My Home Twitza Building , Hitech City Main Rd, Diamond Hills, Hyderabad, Telangana 500081
๐—–๐—ฆ ๐—”๐—น๐—ด๐—ผ ๐Ÿ’ป ๐ŸŒ ใ€Ž๐—–๐—ผ๐—บ๐—ฝ๐—ฒ๐˜๐—ถ๐˜๐—ถ๐˜ƒ๐—ฒ ๐—ฃ๐—ฟ๐—ผ๐—ด๐—ฟ๐—ฎ๐—บ๐—บ๐—ถ๐—ป๐—ดใ€
Photo
#include <bits/stdc++.h>
using namespace std;

const int P = 1000000007;

int solve(string N, int K) {
    if (K == 0) {
        return 1;
    }

    int len = N.size();
    int num[1010] = {0};
    int f[1010][1010][2] = {0};
    int ans = 0;

    for (int i = 2; i <= 1000; i++) {
        num[i] = num[__builtin_popcount(i)] + 1;
    }

    f[0][0][0] = 1;

    for (int i = 0; i < len; i++) {
        for (int j = 0; j <= i; j++) {
            for (int k = 0; k < 2; k++) {
                for (int l = 0; l <= (k ? 1 : (N[i] - '0')); l++) {
                    (f[i + 1][j + l][k | (l < N[i] - '0')] += f[i][j][k]) %= P;
                }
            }
        }
    }

    for (int i = 1; i <= 1000; i++) {
        if (num[i] == K - 1) {
            (ans += f[len][i][0]) %= P;
            (ans += f[len][i][1]) %= P;
        }
    }

    if (K == 1) {
        ans = (ans + P - 1) % P;
    }

    return ans;
}

int main() {
    string s;
    int k;
    cin >> s >> k;
    cout << solve(s, k) << endl;
    return 0;
}


Possible decryption โœ…
๐Ÿ‘2
int solve(int x1, int y1, int x2, int y2, int cx, int cy, int R)
{
    int count = 0;
    int R_squared = R * R;
    for (int y = y1; y <= y2; ++y)
    {
        int dy_squared = (y - cy) * (y - cy);
        if (dy_squared > R_squared)
            continue;
        int dx = static_cast<int>(sqrt(R_squared - dy_squared));
        int min_x = std::max(x1, cx - dx);
        int max_x = std::min(x2, cx + dx);
        if (min_x <= max_x)
        {
            count += (max_x - min_x + 1);
        }
    }

    return count;
}

int main()
{

    int x1 = 1, y1 = 1, x2 = 100000, y2 = 100000, cx = 50000, cy = 50000, R = 1000;
    cin >> x1 >> y1 >> x2 >> y2 >> cx >> cy >> R;
    cout << solve(x1, y1, x2, y2, cx, cy, R) << endl;
    return 0;
}

//Cordinates sprinkler โœ…
#include <bits/stdc++.h>
using namespace std;
int gcd(int a, int b) {
    return b == 0 ? a : gcd(b, a % b);
}
int main() {
    int n, a, b;
    cin >> n >> a >> b;
    map<pair<int, int>, int> slope_map;
    for (int i = 0; i < n; ++i) {
        int x, y;
        cin >> x >> y;
        int dy = y - b;
        int dx = x - a;
        int g = gcd(dx, dy);
        dy /= g;
        dx /= g;
        if (dx < 0) {
            dy = -dy;
            dx = -dx;
        } else if (dx == 0) {
            dy = abs(dy);
        }

        slope_map[{dy, dx}]++;
    }
    int aa = 0;
    for (auto &entry : slope_map) {
        int count = entry.second;
        aa += count * (count - 1) / 2;
    }
    cout << aa << endl;
    return 0;
}


MAQ Software โœ…
๐—–๐—ฆ ๐—”๐—น๐—ด๐—ผ ๐Ÿ’ป ๐ŸŒ ใ€Ž๐—–๐—ผ๐—บ๐—ฝ๐—ฒ๐˜๐—ถ๐˜๐—ถ๐˜ƒ๐—ฒ ๐—ฃ๐—ฟ๐—ผ๐—ด๐—ฟ๐—ฎ๐—บ๐—บ๐—ถ๐—ป๐—ดใ€
Photo
#include <bits/stdc++.h>
using namespace std;

bool isVowel(char c) {
    return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
}

vector<int> hasVowels(vector<string>& strArr, vector<string>& queries) {
    int n = strArr.size();
    vector<int> prefix(n + 1, 0);

    for (int i = 0; i < n; i++) {
        bool startsAndEndsWithVowel = isVowel(strArr[i][0]) && isVowel(strArr[i].back());
        prefix[i + 1] = prefix[i] + (startsAndEndsWithVowel ? 1 : 0);
    }

    vector<int> results;
    for (string query : queries) {
        int l, r;
        sscanf(query.c_str(), "%d-%d", &l, &r);

        l--; r--;

        int count = prefix[r + 1] - prefix[l];
        results.push_back(count);
    }

    return results;
}


IBM HASVOWELโœ…
def lotteryCoupons(n):
    count = [0] * 37
    maxCount = 0

    for i in range(1, n + 1):
        digitSum = sum(int(digit) for digit in str(i))
        count[digitSum] += 1
        maxCount = max(maxCount, count[digitSum])

    result = sum(1 for c in count if c == maxCount)
    return result

IBM Lottery with n coupons โœ…
๐Ÿ”ฅ1
class UnionFind:
    def __init__(self, n):
        self.parent = list(range(n))
        self.rank = [0] * n
        self.components = n

    def find(self, x):
        if self.parent[x] != x:
            self.parent[x] = self.find(self.parent[x])
        return self.parent[x]

    def union(self, x, y):
        px, py = self.find(x), self.find(y)
        if px != py:
            if self.rank[px] < self.rank[py]:
                self.parent[px] = py
            elif self.rank[px] > self.rank[py]:
                self.parent[py] = px
            else:
                self.parent[py] = px
                self.rank[px] += 1
            self.components -= 1

def minOperations(compNodes, compFrom, compTo):
    uf = UnionFind(compNodes + 1)
   
    for u, v in zip(compFrom, compTo):
        uf.union(u, v)
   
    operations = uf.components - 2
    return operations if operations >= 0 else -1

IBM computer networks โœ