HDU 6011

Lotus and Characters

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 ≥0。

Input

First line is T(0≤T≤1000)

denoting the number of test cases.
For each test case,first line is an integer n(1≤n≤26),followed by n lines each containing 2 integers vali,cnti(|vali|,cnti≤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

          

解釋一下測試數據：每次的價值數據都要先進行排序，選出價值大的放在前面

比如第一組測試數據 5 1    6 2；顯然6的價值更高，因此把6放在後面，就是6*2+6*3；再是5，就是5*1，結果就爲35，再是第二組數據：排好序後就變成2  1  -5；

應該就是2*2+1*1=5；如果是2*3+1*2+（-5）*1的話值是小於0的，因此-5數據直接捨棄了。

已經AC過的代碼：

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
pair<int,int> p[35];
int main()
{
    int t,n,v,c;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i=0;i<n;i++)
        {
            scanf("%d %d",&v,&c);
            p[i]=make_pair(v,c);
        }
        long long sum=0;
        long long ans=0;
        sort(p,p+n);
        for(int i=n-1;i>=0;i--)
        {
            for(int j=0;j<p[i].second;j++)
            {
                ans+=p[i].first;
                if(ans<0)
                    break;
                sum+=ans;
            }
        }
        cout<<sum<<'\n';
    }
    return 0;
}
//用到了一些pair函數的使用，此前的博客裏已經說明過了怎麼使用。