# C++ program to find at least how much score it needs to get G amount of score

Suppose we have two arrays p and c both are with D number of elements each, and another number G. Consider in a coding contest, each problem has its score based on difficulty. The problem p[i] has score 100i. These p + ... + p[D] problems are all of the problems present in the contest. a user in the coding site has a number total_score. The total_score of an user is the sum of the following two elements.

• Base score: the sum of score of all solved problems

• Bonus: when a user solves all problems with a score of 100i, he or she earns the perfect bonus c[i] aside the base score.

Amal is new in the contest and has not solved any problem. His objective is to have total score of G or more points. We have to find at least how many problems does he need to solve for this objective.

So, if the input is like G = 500; P = [3, 5]; C = [500, 800], then the output will be 3

## Steps

To solve this, we will follow these steps −

D := size of p
mi := 10000
for initialize i := 0, when i < 1 << D, update (increase i by 1), do:
sum := 0
count := 0
at := 0
an array to store 10 bits b, initialize from bit value of i
for initialize j := 0, when j < D, update (increase j by 1), do:
if jth bit in b is 1, then:
count := p[j]
sum := sum + ((j + 1) * 100 * p[j] + c[j]
Otherwise
at := j
if sum < G, then:
d := (G - sum + (at + 1) * 100 - 1) / ((at + 1) * 100)
if d <= p[at], then:
sum := sum + (at + 1)
count := count + d
if sum >= G, then:
mi := minimum of mi and count
return mi

## Example

Let us see the following implementation to get better understanding −

#include <bits/stdc++.h>
using namespace std;
int solve(int G, vector<int> p, vector<int> c){
int D = p.size();
int mi = 10000;
for (int i = 0; i < 1 << D; i++){
int sum = 0;
int count = 0;
int at = 0;
bitset<10> b(i);
for (int j = 0; j < D; j++){
if (b.test(j)){
count += p.at(j);
sum += (j + 1) * 100 * p.at(j) + c.at(j);
} else {
at = j;
}
}
if (sum < G){
int d = (G - sum + (at + 1) * 100 - 1) / ((at + 1) * 100);
if (d <= p.at(at)){
sum += (at + 1) * 100 * d;
count += d;
}
}
if (sum >= G) {
mi = min(mi, count);
}
}
return mi;
}
int main() {
int G = 500;
vector<int> P = { 3, 5 };
vector<int> C = { 500, 800 };
cout << solve(G, P, C) << endl;
}

## Input

500, { 3, 5 }, { 500, 800 }

## Output

3