hdu----Lotus and Characters

问题描述

Lotus有$n$种字母，给出每种字母的价值以及每种字母的个数限制，她想构造一个任意长度的串。定义串的价值为：第1位字母的价值*1+第2位字母的价值*2+第3位字母的价值*3……求Lotus能构造出的串的最大价值。（可以构造空串，因此答案肯定$\geq 0$）

输入描述

第一行是数据组数$T(0 \leq T \leq 1000)$。对于每组数据，第一行一个整数$n(1 \leq n \leq 26)$，接下来$n$行，每行2个整数$val_i,cnt_i(|val_i|,cnt_i\leq 100)$，分别表示第$i$种字母的价值和个数限制。

输出描述

对于每组数据，输出一行一个整数，表示答案。

输入样例

225 16 23-5 32 11 1

输出样例

355

就是个累加的问题，思维技巧题

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int a[100000];int main(){    int T,n,b,num;    scanf("%d",&T);    while(T--)    {        scanf("%d",&n);        int i = 0,j;        while(n--)        {            scanf("%d %d",&b,&num);            while(num--)                a[i++]=b;        }        sort(a,a+i);        int sum = 0,ans = 0;        for(j = i-1; j >= 0; j--)        {            sum = sum + a[j];            if(sum < 0)                break;            ans += sum;        }        printf("%d\n",ans);    }    return 0;}

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。

 

Inpu

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