Codeforces Round #280 (Div. 2) A B C

http://codeforces.com/contest/492

A 水题

A. Vanya and Cubes

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

Vanya got n cubes. He decided to build a pyramid from them. Vanya wants to build the pyramid as follows: the top level of the pyramid must consist of 1 cube,
the second level must consist of 1 + 2 = 3 cubes, the third level must have 1 + 2 + 3 = 6 cubes,
and so on. Thus, the i-th level of the pyramid must have 1 + 2 + ... + (i - 1) + i cubes.

Vanya wants to know what is the maximum height of the pyramid that he can make using the given cubes.

Input

The first line contains integer n (1 ≤ n ≤ 104)
— the number of cubes given to Vanya.

Output

Print the maximum possible height of the pyramid in the single line.

Sample test(s)

input

1

output

1

input

25

output

4

Note

Illustration to the second sample:

#include <iostream>
#include <algorithm>
#include <iterator>
#include <deque>
#include <vector>
#include <unordered_set>
#include <unordered_map>
#include <set>
#include <valarray>
#include <list>
#include <stack>
#include <array>
#include <iomanip>
#include <map>
#include <string.h>
#include <queue>
#include <sstream>
#include <iomanip>
#include <fstream>
#include <stdio.h>

using namespace std;

int n,m;

int a[1000];

int main ()
{
    a[1] = 1;
    for(int i=1;i<=999;i++)
        a[i] = a[i-1]+i;
    while (cin>>n)
    {
        int i =1;
        int ans = 0;
        while (n-a[i] >=0)
        {
            n-=a[i];
            i++;
            ans++;
        }
        cout<<ans<<endl;
    }
    return 0;
}

B 简单贪心

B. Vanya and Lanterns

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

Vanya walks late at night along a straight street of length l, lit by n lanterns.
Consider the coordinate system with the beginning of the street corresponding to the point 0, and its end corresponding to the point l.
Then the i-th lantern is at the point ai.
The lantern lights all points of the street that are at the distance of at most d from it, where d is
some positive number, common for all lanterns.

Vanya wonders: what is the minimum light radius d should the lanterns have to light the whole street?

Input

The first line contains two integers n, l (1 ≤ n ≤ 1000, 1 ≤ l ≤ 109) —
the number of lanterns and the length of the street respectively.

The next line contains n integers ai (0 ≤ ai ≤ l).
Multiple lanterns can be located at the same point. The lanterns may be located at the ends of the street.

Output

Print the minimum light radius d, needed to light the whole street. The answer will be considered correct if its absolute or relative error doesn't exceed 10 - 9.

Sample test(s)

input

7 15
15 5 3 7 9 14 0

output

2.5000000000

input

2 5
2 5

output

2.0000000000

Note

Consider the second sample. At d = 2 the first lantern will light the segment [0, 4] of
the street, and the second lantern will light segment[3, 5]. Thus, the whole street will be lit.

#include <iostream>
#include <algorithm>
#include <iterator>
#include <deque>
#include <vector>
#include <unordered_set>
#include <unordered_map>
#include <set>
#include <valarray>
#include <list>
#include <stack>
#include <array>
#include <iomanip>
#include <map>
#include <string.h>
#include <queue>
#include <sstream>
#include <iomanip>
#include <fstream>
#include <stdio.h>

using namespace std;

int n,l;
double a[1010];

int main ()
{
    while (cin>>n>>l)
    {
          for(int i=1;i<=n;i++)
                cin>>a[i];
          sort(a+1,a+1+n);

          double ans = a[1] ;

          for(int i=1;i<=n-1;i++)
          {
              ans = max (ans,(a[i+1]-a[i])*1.0/2);
          }

          ans = max(ans,l-a
);

          printf("%.10lf\n",ans);
    }
    return 0;
}

C 按照bi升序排列 注意不要超时即可

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 45 2
4 7
3 13 2
2 5

output

4

input

2 5 45 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.

#include <iostream>
#include <algorithm>
#include <iterator>
#include <deque>
#include <vector>
#include <unordered_set>
#include <unordered_map>
#include <set>
#include <valarray>
#include <list>
#include <stack>
#include <array>
#include <iomanip>
#include <map>
#include <string.h>
#include <queue>
#include <sstream>
#include <iomanip>
#include <fstream>
#include <stdio.h>

using namespace std;

int n;
long long r;
long long avg,***E;

struct q
{
    int a;
    int b;
}p[100010];

bool cmp(q aa,q bb)
{
    return aa.b < bb.b;
}

int main ()
{
    while (cin>>n>>r>>avg)
    {
        ***E= 0;
        for(int i=1;i<=n;i++)
        {
            cin>>p[i].a>>p[i].b;
            ***E+=p[i].a;
        }

        if (***E >= avg*n)
        {
            cout<<0<<endl;
            return 0;
        }

        sort(p+1,p+1+n,cmp);

        long long  ans = 0;

        int i = 1;
        while ( ***E < avg*n )
        {
            ans += p[i].b * min ( r -p[i].a,n*avg - ***E );
            
            ***E += min(r -p[i].a,n*avg - ***E);

            i++;
        }

        cout<<ans<<endl;
                    
    }

    return 0;
}