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

225 16 23-5 32 11 1

 

Sample Output

355

 

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;elsecout<<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;}