Yogurt factory （贪心）

Yogurt factory

The cows have purchased a yogurt factory that makes world-famous Yucky Yogurt. Over the next N (1 <= N <= 10,000) weeks, the price of milk and labor will fluctuate weekly such that it will cost the company C_i (1 <= C_i <= 5,000) cents to produce one unit of yogurt in week i. Yucky's factory, being well-designed, can produce arbitrarily many units of yogurt each week. 

Yucky Yogurt owns a warehouse that can store unused yogurt at a constant fee of S (1 <= S <= 100) cents per unit of yogurt per week. Fortuitously, yogurt does not spoil. Yucky Yogurt's warehouse is enormous, so it can hold arbitrarily many units of yogurt. 

Yucky wants to find a way to make weekly deliveries of Y_i (0 <= Y_i <= 10,000) units of yogurt to its clientele (Y_i is the delivery quantity in week i). Help Yucky minimize its costs over the entire N-week period. Yogurt produced in week i, as well as any yogurt already in storage, can be used to meet Yucky's demand for that week.

Input

* Line 1: Two space-separated integers, N and S. 

* Lines 2..N+1: Line i+1 contains two space-separated integers: C_i and Y_i.

Output

* Line 1: Line 1 contains a single integer: the minimum total cost to satisfy the yogurt schedule. Note that the total might be too large for a 32-bit integer.

Sample Input

4 5
88 200
89 400
97 300
91 500

Sample Output

126900

Hint

OUTPUT DETAILS: 
In week 1, produce 200 units of yogurt and deliver all of it. In week 2, produce 700 units: deliver 400 units while storing 300 units. In week 3, deliver the 300 units that were stored. In week 4, produce and deliver 500 units. 

题目大意：有一个酸奶工厂，他要向客户供奶，一共有n周，但是生产酸奶的单位价格每一周都不一样，他有一个仓库，可以保存酸奶，每保存一周有固定的保存费用s，现在给出n周的每周生产单位酸奶的价格和每周要和客户供奶的量，求最小花费是多少

代码：
 

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct node
{
    long long c,y,d;
}a[10005];
int main()
{
   long long n,m;
   while(~scanf("%lld%lld",&n,&m))
   {
       for(int i=0;i<n;i++)
        scanf("%lld%lld",&a[i].c,&a[i].y),a[i].d=i;
       long long ans=a[0].c*a[0].y,k=0,xa,xb;
       for(int i=1;i<n;i++)
       {
           xa=a[i].c*a[i].y,xb=(a[k].c+m*(i-k))*a[i].y;
           if(xa>xb)
               ans+=xb;
           else
           {
               ans+=xa;
               k=i;
           }
       }
       printf("%lld\n",ans);
   }
   return 0;
}