Pearls（动态规划）

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

 

题目意思是说，有几种不同的珍珠。每种珍珠都有它的单价。当然质量高的珍珠价格一定也是高的。
为了避免买家只买1个珍珠。就要求不论是买了多少个珍珠都是需要在购买数量上加10.之后乘上单价。
求出总的花费！例如：买5个单价是10的珍珠。需要的花费是（5+10）*10= 150.买100个单价是20的珍珠
需要的花费是（100+10）*20= 2200.总共需要的花费是150+2200=2350.如果把珍珠的质量提高了。需要的105个
珍珠都买单价是20的。也就是说都买质量好的。总的花费是（5+100+10）*20= 2300.在两组数据看来。珍珠都
买了高品质的了，而且花费也少了！
问题是怎么样能花费最少买珍珠！

如果只有两类zz即分为1：完全用第二种，2：两种都用
如果有三类 zz  分为1：完全用第三类，2：用第三类和第二类；3：用第一二三类；
如果有四类 zz  分为1：完全用第四类，2：用第三类类与第四类；3：用第四三二类；4：用第四三二一类；
每一类都要比较一下判断最优解记录在一个数组中ans[]；这里有子问题结构，用第二类和第一类此时第二类要替换第一类，与只有一二类时完全用第二类的问题具有重叠可以dp即直接加上ans[2];在比较；

#include<stdio.h>
#include<string.h>
int min(int n,int m)
{
    return n<m?n:m;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,i,a[105],p[105],j,b;
        int dp[1000],sum[1000];
        scanf("%d",&n);
        memset(sum,0,sizeof(sum));
        for(i=1;i<=n;i++)
        {
            scanf("%d%d",&a[i],&p[i]);
            sum[i]=sum[i-1]+a[i];//sum[i]代表前i个种类有多少珠宝；
        }
        memset(dp,0,sizeof(dp));
        for(i=1;i<=n;i++)
        {
            dp[i]=(a[i]+10)*p[i]+dp[i-1];
            for(j=0;j<=i;j++)//一定从0开始包含所有情况；
            dp[i]=min(dp[i],(dp[j]+(sum[i]-sum[j]+10)*p[i]));
        }
        printf("%d\n",dp[n]);
    }
    return 0;
}