Codeforces Round #280 (Div. 2) C. Vanya and Exams 贪心

C. Vanya and Exams

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/492/problem/C

Description

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 Input

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

Sample Output

4

HINT

题意

 有一个人有n门课程，每一门课程他最多获得r学分，他只要所有课程的平均学分有avg，他就可以获得奖学金

每门课程，他已经获得了ai学分，剩下的每一个学分，都需要写bi篇论文才能得到

然后问你，这个人最少写多少论文才能获得奖学金

题解：

贪心，我们选择bi最小的开始写论文，然后扫一遍就好了，直到学分够为止

代码

#include<iostream>
#include<stdio.h>
#include<algorithm>
using namespace std;

#define maxn 100005
pair<long long,long long> p[maxn];
int main()
{
    int n;
    long long r,avg;
    scanf("%d%lld%lld",&n,&r,&avg);
    avg*=n;
    for(int i=0;i<n;i++)
    {
        scanf("%lld%lld",&p[i].second,&p[i].first);
        avg-=p[i].second;
        p[i].second = r - p[i].second;
    }
    sort(p,p+n);
    long long ans = 0;
    for(int i=0;i<n;i++)
    {
        if(avg<=0)break;
        if(p[i].second>=avg)
        {
            ans+=avg*p[i].first;
            break;
        }
        else
        {
            ans+=p[i].second*p[i].first;
            avg-=p[i].second;
        }
    }
    printf("%lld
",ans);
}