问题描述：
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.
输入说明：
-
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.
输出说明：
- 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
SAMPLEOUTPUT：
126900
hints：
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.
思路：
题意，每周我们需要交付一定量的酸奶，但是可以生产无限量的酸奶，储存酸奶需要花费一定的费用，每周生产的cost都是不一样的，我们应该如何生产能够让成本最少。一道很经典的贪心问题，首先第一周至少要生产所需要的量，其次，如果前一周要生产之后的酸奶，需要比较的是当周的生产费用+储存费用与下周的生产费用的大小关系。因此在处理时，依次输入n周的生产所需和所需的酸奶量，用一个变量储存当前最划算的生产方式，每周都进行比较，如果之前某周的生产更划算，就在那一周多生产当前周所需的酸奶数，反之更新变量的储存然后按照当前周的计划进行生产。
AC代码：
#include<iostream>
using namespace std;
int main()
{
int n,s,c=0,ci,yi;
long long ans=0;
cin>>n>>s;
for(int i=1;i<=n;i++)
{
cin>>ci>>yi;
if(i==1)
{
ans+=ci*yi;
c=ci+s;
}
else
{
if (c>ci)
{
ci=ci;
}
else
{
ci=c;
}
ans+=ci*yi;
c=ci+s;
}
}
cout<<ans<<endl;
return 0;
}