Lotus and Characters

      HDU 6011

Description

Lotus has $n$ kinds of characters,each kind of characters has a value and a amount.She wants to construct a string using some of these characters.Define the value of a string is:its first character's value*1+its second character's value *2+...She wants to calculate the maximum value of string she can construct.
Since it's valid to construct an empty string,the answer is always $\geq 0$。

 

Input

First line is $T(0 \leq T \leq 1000)$ denoting the number of test cases.
For each test case,first line is an integer $n(1 \leq n \leq 26)$,followed by $n$ lines each containing 2 integers $val_i,cnt_i(|val_i|,cnt_i\leq 100)$,denoting the value and the amount of the ith character.

 

Output

For each test case.output one line containing a single integer,denoting the answer.

 

Sample Input

     

       2
2
5 1
6 2
3
-5 3
2 1
1 1 
     

 

Sample Output

     

       35
5 
     
     

      
#include<stdio.h>
#include<math.h>
#include<iostream>
#include<string.h>
#include<algorithm>
using namespace std;
int main()
{
    int T,n,i,j,sum,b[10001],h,l,ans,k;
    while(scanf("%d",&T)!=EOF)
    {
        while(T--)
        {sum=0;j=0;
            scanf("%d",&n);
            for(i=0;i<n;i++)
            {
                scanf("%d%d",&h,&l);
                while(l--)
                {
                    b[j]=h;
                    j++;
                }
            }
            sort(b,b+j);
            ans=0;
            for(k=0;;k++)
            {
                sum=0;
            for(i=k;i<j;i++)
            {
                sum=sum+b[i]*(i+1-k);
            }
            if(sum>ans)
            {ans=sum;}
            if(b[k]>0)
            {
                break;
            }
            }

            printf("%d\n",ans);
        }

    }
    return 0;
}