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

这一题的dp比较好理解，关键是从英文转化成中文再转化成目标性的语句有困难。

题意：

　　先给t组测试样例，再给出n组数据，代表n组珍珠质量等级数，前者代表珍珠所属的质量等级，后者代表该质量等级里面的每颗珍珠的价格，质量等级和珍珠的价格都是升序。第一组数据说的是如果我要买3颗珍珠，一个等级一个等级购买的话需要(100+10)*1+(100+10)*2 =330元，但是如果全部按第二等级来买，需要(202+10)*2=424元，所以最小花费是330元，以此类推。要求求出最少花费价格去购买所有数量的珍珠。

https://blog.csdn.net/SDUTyangkun/article/details/52225285  这里面讲的很详细了

自己的理解：

//解该题的关键在于珍珠的类别（也就是质量）和数量是呈上升趋势的
//每一个状态都是从上一个状态推算来的，所以可以推出动态规划的表达式

#include
#include
#include
#include
#include
#include

struct node
{
int num;
int p;
}a[];

int dp[],sum[];
int main()
{
std::ios::sync_with_stdio(false);
cin.tie();
cout.tie();
int t,n;
cin>>t;
while(t--)
{
memset(dp,,sizeof(dp));
memset(a,,sizeof(a));
memset(sum,,sizeof(sum));
cin>>n;
sum[]=;
for(int i=;i<=n;i++) { cin>>a[i].num>>a[i].p;
sum[i]=sum[i-]+a[i].num;
}

     dp\[\]=;  
     for(int i=;i<=n;i++)  
     {  
         dp\[i\]=(a\[i\].num+)\*a\[i\].p+dp\[i-\];  
         //需要加上dp\[i-1\]  
         //因为(a\[i\].num+10)\*a\[i\].p是当前位置的珍珠的价格，还未进行优化之前的；  
         //需要再加上之前所有状态的  
         for(int j=;j<i;j++)  
             //注意一下，这里的j从0开始，不是从1  
             //因为是从上一个状态得来的，要是从1开始，在i=1的时候会产生错误  
         {  
             dp\[i\]=min(dp\[i\],dp\[j\]+(sum\[i\]-sum\[j\]+)\*a\[i\].p);  
         }  
     }  
     cout<<dp\[n\]<<endl;  
 }  
 return ;  

}