poj 1260

Description

In Pearlania everybody is fond of pearls. One company, called The Royal Pearl, produces a lot of jewelry with pearls in it. The Royal Pearl has its name because it delivers to the royal family of Pearlania. But it also produces bracelets and necklaces for ordinary
people. Of course the quality of the pearls for these people is much lower then the quality of pearls for the royal family.In Pearlania pearls are separated into 100 different quality classes. A quality class is identified by the price for one single pearl
in that quality class. This price is unique for that quality class and the price is always higher then the price for a pearl in a lower quality class.

Every month the stock manager of The Royal Pearl prepares a list with the number of pearls needed in each quality class. The pearls are bought on the local pearl market. Each quality class has its own price per pearl, but for every complete deal in a certain
quality class one has to pay an extra amount of money equal to ten pearls in that class. This is to prevent tourists from buying just one pearl.

Also The Royal Pearl is suffering from the slow-down of the global economy. Therefore the company needs to be more efficient. The CFO (chief financial officer) has discovered that he can sometimes save money by buying pearls in a higher quality class than is
actually needed.No customer will blame The Royal Pearl for putting better pearls in the bracelets, as long as the

prices remain the same.

For example 5 pearls are needed in the 10 Euro category and 100 pearls are needed in the 20 Euro category. That will normally cost: (5+10)*10+(100+10)*20 = 2350 Euro.Buying all 105 pearls in the 20 Euro category only costs: (5+100+10)*20 = 2300 Euro.

The problem is that it requires a lot of computing work before the CFO knows how many pearls can best be bought in a higher quality class. You are asked to help The Royal Pearl with a computer program.

Given a list with the number of pearls and the price per pearl in different quality classes, give the lowest possible price needed to buy everything on the list. Pearls can be bought in the requested,or in a higher quality class, but not in a lower one.

Input

The first line of the input contains the number of test cases. Each test case starts with a line containing the number of categories c (1<=c<=100). Then, c lines follow, each with two numbers ai and pi. The first of these numbers is the number of pearls ai
needed in a class (1 <= ai <= 1000).

The second number is the price per pearl pi in that class (1 <= pi <= 1000). The qualities of the classes (and so the prices) are given in ascending order. All numbers in the input are integers.

Output

For each test case a single line containing a single number: the lowest possible price needed to buy everything on the list.

Sample Input

2

2

100 1

100 2

3

1 10

1 11

100 12

Sample Output

330 1344

题目大意就是，一家珠宝公司要根据需求去买珍珠，有许多不同质量的珍珠，高质量的珍珠比低质量的要贵，可以用高质量的珍珠来代替低质量的珍珠，有一个规定是买一种质量珍珠就要多付十个该种珍珠的钱，求出最小费用。

用高质量的珍珠替代低质量的珍珠时，这两个应该是连续的，因为如果有某种质量的珍珠介于这两者之间，那么用中间的那个来替代低质量的珍珠所花的代价一定要比高质量的要小。所以某一质量的珍珠可以用来替代该质量之前连续的几种质量的珍珠，用ans[i]表示到前i中质量珍珠的最小费用，s[i]表示前i中质量珍珠的总数量，有方程：

ans[i] = min(ans[i], ans[j] + (s[i] - s[j] + 10) * p[i])

#include <iostream>
#include <cstdio>
#include <climits>

#define min(a, b) (a > b ? b : a)

int main()
{
int i, j, n, c, t, a[101], p[101], s[101], ans[101];

std::cin >> n;
while (n--) {
std::cin >> c;
s[0] = a[0];
for (i = 1; i != c + 1; ++i) {
scanf("%d%d", &a[i], &p[i]);
s[i] = s[i - 1] + a[i];
}
ans[0] = 0;
for (i = 1; i != c + 1; ++i) {
ans[i] = INT_MAX;
for (j = 0; j != i; ++j) {
t = ans[j] + (s[i] - s[j] + 10) * p[i];
ans[i] = min(ans[i], t);
}
}
std::cout << ans[c] << std::endl;
}

return 0;
}