cf492C 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.

唉其实就是贪心完排个序

结果没有加个(LL)就被x掉了

不开心

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<queue>
#include<deque>
#include<set>
#include<map>
#include<ctime>
#define LL long long
#define inf 0x7ffffff
#define pa pair<int,int>
#define pi 3.1415926535897932384626433832795028841971
using namespace std;
inline LL read()
{
    LL x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n,m,k;
LL tot,ans;
struct cla{
	int a,b;
}c[1000010];
bool operator <(cla a,cla b)
{return a.b<b.b;}
int main()
{
	n=read();m=read();k=read();
	for (int i=1;i<=n;i++)
	{

		c[i].a=read();c[i].b=read();
		tot+=c[i].a;
		c[i].a=m-c[i].a;
	}
	tot=(LL)n*k-tot;
	if (tot<=0)
	{
		printf("0\n");
		return 0;
	}
	sort(c+1,c+n+1);
	for (int i=1;i<=n;i++)
	{
		if (!tot)break;
		if (c[i].a<=tot)
		{
			tot-=c[i].a;
			ans+=(LL)c[i].b*c[i].a;
		}else
		if (c[i].a>tot)
		{
			ans+=(LL)tot*c[i].b;
			tot=0;
		}
	}
	printf("%lld\n",ans);
}