HDU 6011

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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

225 16 23-5 32 11 1

Sample Output

解释一下测试数据：每次的价值数据都要先进行排序，选出价值大的放在前面

比如第一组测试数据 5 1 6 2；显然6的价值更高，因此把6放在后面，就是6*2+6*3；再是5，就是5*1，结果就为35，再是第二组数据：排好序后就变成2 1 -5；

应该就是2*2+1*1=5；如果是2*3+1*2+（-5）*1的话值是小于0的，因此-5数据直接舍弃了。

已经AC过的代码：

#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函数的使用，此前的博客里已经说明过了怎么使用。

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