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

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

Note

In the first sample Vanya can write 2 essays for the 3rd exam to raise his grade by 2 points and 2 essays for the 4th exam to raise his grade by 1 point.

In the second sample, Vanya doesn't need to write any essays as his general point average already is above average.

<pre name="code" class="cpp" style="color: rgb(34, 34, 34);">/*题目大意：Vanya想获得奖学金，条件是她所有科目的平均分要达到avg，当然没有达到也是没有关系滴、可以通过
 *			写文章来加分、这就好比每年综合测评时候有的同学文化成绩不行还可以通过这加分那加分来补一样、相应科目加上一分
 *			要写多少文章告诉你，让你求出最少写多少文章能够获得奖学金
 *输入：n，代表科目数量、r，代表 每门课最高达到的分数，avg，表示平均分。接下来的n行表示第i门课的分数与在该门课
 *		上面加上一分所要写出的文章数量 
 *输出：最少写的文章数量 
 *算法分析：采用贪心策略，写文章肯定是写那种写得数量少加上一分的、 
*/
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <algorithm>
#include <stack>
using namespace std;

typedef long long int LL ;

struct node {
	LL x, y;
}a[100010];

int cmp(node c, node b) {
	return c.y < b.y;
}

int main() {
	memset(a, 0, sizeof(a));

	LL n, r, avg;
	cin >> n >> r >> avg;
	for (LL i = 0; i<n; i++)
		cin >> a[i].x >> a[i].y;
	LL res = n *avg;
	LL sum = 0;
	for (LL i = 0; i<n; i++)
		sum += a[i].x;
	LL p = res - sum;
	if (p > 0) {
		res = 0;
		sort(a, a+n, cmp);
		LL f = 0, flag1 = 1;
		for (LL i = 0; i<n && flag1; i++) {
			if (a[i].x < r && f < p) {
				int flag = r-a[i].x;
				f += flag;
				if (f <= p) {
					res += flag*a[i].y;
				}
				else {
					f -= flag;
					int p1 = p-f;
					res += a[i].y*p1;
					flag1 = 0;
					break;
				}
				
				
			}
		}
		cout << res << endl;
	}
	else 
		cout << 0<< endl;
	return 0;
}

没有AC不了的题，只有不努力的ACMer。