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C++ code to find winner of math contest
Suppose we have two arrays P and T of size n. And have another number c. Amal and Bimal are going to participate one math contest. There are n problems. The ith problem has initial score P[i], and takes T[i] to solve it. P and T both are sorted in increasing order. Here c is the constant for loosing points. If a problem is submitted at time x (x minutes after starting the contest), it gives max(0, P[i] - c*x) points. Amal is going to solve problems in order 1, 2, ... n and Bimal is going to solve them like n, n-1, ... 1. We have to find who will get the maximum score. If they have got the same, it will be a 'Tie'.
So, if the input is like c = 2; P = [50, 85, 250]; T = [10, 15, 25], then the output will be Amal.
Steps
To solve this, we will follow these steps −
n := size of P m := 0 ans1 := 0 ans2 := 0 for initialize i := 0, when i < n, update (increase i by 1), do: m := m + T[i] ans1 := ans1 + maximum of (0, P[i] - c * m) m := 0 for initialize i := n - 1, when i > 1, update (decrease i by 1), do: m := m + T[i] ans2 := ans2 + maximum of (0, P[i] - c * m) if ans1 > ans2, then: return "Amal" otherwise when ans1 < ans2, then: return "Bimal" Otherwise return "Tie"
Example
Let us see the following implementation to get better understanding −
#include <bits/stdc++.h>
using namespace std;
string solve(int c, vector<int> P, vector<int> T){
int n = P.size();
int m = 0, ans1 = 0, ans2 = 0;
for (int i = 0; i < n; i++){
m += T[i];
ans1 += max(0, P[i] - c * m);
}
m = 0;
for (int i = n - 1; i > 1; i--){
m += T[i];
ans2 += max(0, P[i] - c * m);
}
if (ans1 > ans2)
return "Amal";
else if (ans1 < ans2)
return "Bimal";
else
return "Tie";
}
int main(){
int c = 2;
vector<int> P = { 50, 85, 250 };
vector<int> T = { 10, 15, 25 };
cout << solve(c, P, T) << endl;
}Input
2, { 50, 85, 250 }, { 10, 15, 25 }Output
Amal
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