CodeForces 492C Vanya and Exams （贪心）

C. Vanya and Exams

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Vanya wants to pass n exams and get the academic scholarship. He will get the scholarship if the average grade mark for all the exams is at least avg. The exam grade cannot exceed r. Vanya has passed the exams and got grade ai for the i-th exam. To increase the grade for the i-th exam by 1 point, Vanya must write bi essays. He can raise the exam grade multiple times.

What is the minimum number of essays that Vanya needs to write to get scholarship?

Input

The first line contains three integers n, r, avg (1 ≤ n ≤ 105, 1 ≤ r ≤ 109, 1 ≤ avg ≤ min(r, 106)) — the number of exams, the maximum grade and the required grade point average, respectively.

Each of the following n lines contains space-separated integers ai and bi (1 ≤ ai ≤ r, 1 ≤ bi ≤ 106).

Output

In the first line print the minimum number of essays.

Sample test(s)

input

5 5 4
5 2
4 7
3 1
3 2
2 5

output

4

input

2 5 4
5 2
5 2

output

0

思路： 平均分其实乘一下n以后就是总分，也就是说总分要达到这个sum=avg*n，而每一科提升一分的代价是bi个论文，提升空间则还有r-ai个学分。计算好每个科目以后，贪心取代价最小的，按bi从小到大排序，能取多少就取多少，直到分数达到sum为止，性价比最高的科目的提分空间用完以后再去性价比第二高的科目。

#include <bits/stdc++.h>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef pair<ll,ll> pii;
const int INF = 1e9;
const double eps = 1e-;
//const int N = ;
int cas = ;

ll n,r,avg,sum;

void run()
{
    ll a,b;
    vector<pii> v;
    sum = avg * n;
    for(int i = ; i < n; i++ )
    {
        cin >> a >> b;
        sum -= a;
        v.push_back(make_pair(b,r-a));
    }
    sort(v.begin(),v.end());
    ll ans = ;
    for(int i = ; i < n; i++ )
    {
        if(sum <= ) break;
        if(sum >= v[i].second)
            ans+=v[i].first*v[i].second, sum-=v[i].second;
        else
            ans += sum*v[i].first, sum -= sum;
    }
    cout << ans << endl;
}

int main()
{
    #ifdef LOCAL
//    freopen("case.txt","r",stdin);
    #endif
    while(cin >> n >> r >> avg)
        run();
    return ;
}