hdu 6011 Lotus and Characters

Lotus has nn
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≥0。
InputFirst line is T(0≤T≤1000)T(0≤T≤1000)
denoting the number of test cases.

For each test case,first line is an integer
n(1≤n≤26)n(1≤n≤26),followed
by nn
lines each containing 2 integers
vali,cnti(|vali|,cnti≤100)vali,cnti(|vali|,cnti≤100),denoting
the value and the amount of the ith character.
OutputFor 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

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<iostream>
#include<set>

using namespace std;

#define  mem(x,y) memset(x,y,sizeof(x))
struct node
{
int num, value;
int f=0;
friend bool operator<(const node &a, const node &b)
{
return a.value < b.value;
}
};
node zi[10000];
int n;
int jia(int i)
{
int t = 0;
for (int j = 1; j <= i; j++)
{
t += j;
}
return t;
}
int main()
{

int t;
scanf("%d", &t);
while (t--)
{
scanf("%d", &n);
int sum=0;
for (int i = 0; i < n; i++)
{
scanf("%d %d", &zi[i].value, &zi[i].num);
if (zi[i].value > 0)
{
sum += zi[i].num*zi[i].value;
}
}

sort(zi, zi + n);
int a;
int ans=0;
int temp = 0;
for (int i = n-1,k=1; i >= 0; i--)
{
int nowans = ans;
for (int j = 0; j < zi[i].num; j++)
{
nowans = temp + nowans + zi[i].value;
if (nowans > ans)
{
ans = nowans;
a = zi[i].value;
temp += a;
}
else if (nowans < ans)
{
break;
}
}
}
printf("%d\n", ans);

}
return 0;
}

//by swust_t_p