hdu 6011Lotus and Characters

Lotus and Characters

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/131072 K (Java/Others)
Total Submission(s): 1199    Accepted Submission(s): 409

Problem 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 ≥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

 

Source

BestCoder Round #91

 

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#include <iostream>
#include<math.h>
#include<algorithm>
#include<memory.h>
using namespace std;
long long int a[7000000];
long long int Maxn(long long int n,long long int a[])
{
	long long int max1=-1000000,t=1,sum;
	for(int i=1;i<=n;i++)
	{
		sum=0;
		t=1;
	 	for(int j=n-i;j<n;j++)//到着从最后开始数，假如只有一个数，取最后一个，假如只有两个数，取最后一个和倒数第二个 
	 	{	
	 	  	sum=sum+t*a[j]; 	  
	 	  	t++;	 
		}
		if(max1<sum)  max1=sum;
		if(sum<0) break;
		//Since it's valid to construct an empty string,the answer is always ≥0。
		//这句是说，如果输入的全是负数的话，输出0,坑死啦。。。	
	}
	if(max1<0) cout<<0<<endl;
	else	cout<<max1<<endl;	
}
int main(void)
{
   int num;
   cin>>num;
   while(num--)
   {
   		long long int n,k=0,x,y;
   		cin>>n;
   		for(int i=0;i<n;i++)
   		{  
		   cin>>x>>y;
		   while(y--)
		   {
		   	 a[k++]=x;	
		   }  
		}
		sort(a,a+k);//从小到大排序 
		Maxn(k,a);
   }
   return 0;
}