「CF492C」Vanya and Exams

2014年12月2日3,4172

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 a_i for the i-th exam. To increase the grade for the i-th exam by 1 point, Vanya must write b_i 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 a_i and b_i (1 ≤ a_i ≤ r, 1 ≤ b_i ≤ 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

2 5 4

5 2

output

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.

题解

贪心选取b最小的。。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<map>
#include<ctime>
#include<vector>
#include<set>
#include<cmath>
#include<algorithm>
#define inf 1000000000
#define ll long long
#define pa pair<int,int>
using namespace std;
int read()
{
    int 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,r;
ll avg,sum,lim,ans;
struct data{int a,b;ll lim;}a[100005];
bool operator<(data a,data b)
{
    return a.b<b.b;
}
int main()
{
    n=read();r=read();avg=read();
    avg*=n;
    for(int i=1;i<=n;i++)
    {
        a[i].a=read(),a[i].b=read(),sum+=a[i].a;
        a[i].lim=(r-a[i].a);
    }
    if(sum>=avg){puts("0");return 0;}
    sort(a+1,a+n+1);
    for(int i=1;i<=n&&sum<avg;i++)
    {
        ll t=min(avg-sum,a[i].lim);
        sum+=t;
        ans+=t*a[i].b;
    }
    printf("%I64d",ans);
    return 0;
}

#include<iostream>

#include<cstdio>

#include<cstring>

#include<cstdlib>

#include<map>

#include<ctime>

#include<vector>

#include<set>

#include<cmath>

#include<algorithm>

#define inf 1000000000

#define ll long long

#define pa pair<int,int>

using namespace std;

int read()

{

int 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,r;

ll avg,sum,lim,ans;

struct data{int a,b;ll lim;}a[100005];

bool operator<(data a,data b)

{

return a.b<b.b;

}

int main()

{

n=read();r=read();avg=read();

avg*=n;

for(int i=1;i<=n;i++)

{

a[i].a=read(),a[i].b=read(),sum+=a[i].a;

a[i].lim=(r-a[i].a);

}

if(sum>=avg){puts("0");return 0;}

sort(a+1,a+n+1);

for(int i=1;i<=n&&sum<avg;i++)

{

ll t=min(avg-sum,a[i].lim);

sum+=t;

ans+=t*a[i].b;

}

printf("%I64d",ans);

return 0;

}

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