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#include <iostream> #include <string> #include <iomanip> #include <vector> #include <algorithm> struct Client { std::string name; double total_invested_in_bonds; double total_invested_in_stocks; }; bool compareByBonds(const Client &amp;a, const Client &amp;b) { return a.total_invested_in_bonds &gt; b.total_invested_in_bonds; } int main() { std::vector<client> clients = { {"Client1", 7500.0, 3000.0}, {"Client2", 6000.0, 5000.0}, {"Client3", 4000.0, 6000.0}, // Add more clients as needed }; // Sort clients based on total_invested_in_bonds in descending order std::sort(clients.begin(), clients.end(), compareByBonds); // Display the result std::cout &lt;&lt; std::left &lt;&lt; std::setw(15) &lt;&lt; "Client Name" &lt;&lt; std::setw(25) &lt;&lt; "Total Invested in Bonds" &lt;&lt; "Total Invested in Stocks" &lt;&lt; std::endl; for (const auto &amp;client : clients) { if (client.total_invested_in_bonds &gt; 5000.00) { std::cout &lt;&lt; std::left &lt;&lt; std::setw(15) &lt;&lt; client.name &lt;&lt; std::fixed &lt;&lt; std::setprecision(2) &lt;&lt; std::setw(25) &lt;&lt; client.total_invested_in_bonds &lt;&lt; client.total_invested_in_stocks &lt;&lt; std::endl; } } return 0; } You can modify the clients vector to include more clients with their respective investments in bonds and stocks. When you run this program, it will display the result according to the specified requirements.</client></algorithm></vector></iomanip></string></iostream> Telegram:- @allcoding1_official

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8:15

To find the ratio of A to C, we can use the given ratios for AB and BC. Given: AB = 2:3 BC = 4:5 We want to find the ratio of A to C. One way to solve this is to express both AB and BC in terms of a common variable and then use that to find the ratio of A to C. Let's express AB and BC in terms of a common multiple of B. Let's assume the common multiple is x. Then: AB = 2x : 3x BC = 4x : 5x Now, to find the ratio of A to C, we can multiply the two ratios together: A:B * B:C = A:C So: (2x : 3x) * (4x : 5x) = (2x * 4x : 3x * 5x) = (8x^2 : 15x^2) Therefore, the ratio of (A to C is 8:15)

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int min(string str){ unordered_map<char,int>mp; for(char :str){   mp[ch]++; } unodered_set<int>d; for(auto it:mp){   d.insert(it.second); } return d.size(); } Music Teacher Only 4 test cases pass Telegram:- @allcoding1_official

#include <iostream> #include <cmath> struct Circle { double x; // x-coordinate of the center double y; // y-coordinate of the center double r; // radius }; double distanceBetweenCenters(Circle A, Circle B) { return sqrt(pow((B.x - A.x), 2) + pow((B.y - A.y), 2)); } int main() { Circle A, B; // Example values A.x = 0; A.y = 0; A.r = 5; B.x = 10; B.y = 0; B.r = 7; double distance = distanceBetweenCenters(A, B); double sumOfRadii = A.r + B.r; double differenceOfRadii = abs(A.r - B.r); if (distance == sumOfRadii) { std::cout &lt;&lt; "The circles are touching at a single point." &lt;&lt; std::endl; } else if (distance &lt; sumOfRadii) { std::cout &lt;&lt; "The circles are intersecting." &lt;&lt; std::endl; } else if (distance == differenceOfRadii) { std::cout &lt;&lt; "The circles are touching from within or without." &lt;&lt; std::endl; } else { std::cout &lt;&lt; "The circles are not intersecting." &lt;&lt; std::endl; } if ((A.x == B.x) &amp;&amp; (A.y == B.y) &amp;&amp; (A.r == B.r)) { std::cout &lt;&lt; "The circles are concentric." &lt;&lt; std::endl; } return 0; } C++ Telegram:- @allcoding1_official

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#include<bits/stdc++.h> using namespace std; int main(){     int n,m;     cin>>n>>m;     vector<int>ans(n);     if(n<=m){         for(int i=0;i<n;i++){             ans[i]=i+1;         }     }     else{                  int k=n/m;         // int r=n%m;         int j=0;         for(int i=0;i<n;i++){             ans[i]=j+1;             j++;             j%=m;         }                       }     for(int i=0;i<n;i++){         cout<<ans[i]<<" ";     }     cout<<endl; } King Dreams ✅ Intuit Telegram:- @meterials_available

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class Solution { public:   int minOperations(vector& nums, int x, int y) {     int l = 0;     int r = ranges::max(nums);     while (l < r) {       const int m = (l + r) / 2;       if (isPossible(nums, x, y, m))         r = m;       else         l = m + 1;     }     return l;   } private:   bool isPossible(const vector& nums, int x, int y, int m) {     long long additionalOps = 0;     for (const int num : nums)       additionalOps += max(0LL, (num - 1LL * y * m + x - y - 1) / (x - y));     return additionalOps <= m;   } }; Job Execution ✅