Hdu 1300 Pearls (dp)

Hdu 1300 Pearls (dp)

Question:

A variety of pearls, each purchase must add 10 to the original quantity.

For example, buy 5 pearls with a unit price of 10. The cost is (5 + 10) * 10 = 150. buy 100 pearls with a unit price of 20.
The cost is (100 + 10) * 20 = 2200. The total cost is 150 + 2200 = 2350. if the quality of the pearl is improved. Required 105
The price of pearls is 20. That is to say, both products are of good quality. The total cost is (5 + 100 + 10) * 20 = 2300. In view of the two groups of data. Pearl caPital
I bought high-quality products, and the cost is low!
The question is how to buy at least Pearl!

Train of Thought Analysis:

Dp [I] indicates the optimal solution for buying pearls in I.

Status transfer: dp [I] = min (dp [j] + v) v indicates that the pearls between j-I are all purchased at the price of I.

#include 
 
  #include 
  
   #include 
   
    #include #define maxn 105using namespace std;struct node{    int price,num;}a[105];int dp[maxn];int sum[maxn];int main(){    int T;    scanf("%d",&T);    while(T--)    {        int n;        scanf("%d",&n);        for(int i=1;i<=n;i++)        {            scanf("%d%d",&a[i].num,&a[i].price);            sum[i]=sum[i-1]+a[i].num;        }        memset(dp,0x3f,sizeof dp);        dp[0]=0;        for(int i=1;i<=n;i++)        {            for(int j=0;j