SDAU课程练习1015

Problem Description

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

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

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 isthe delivery quantity in week i). Help Yucky minimize its costsover the entire N-week period. Yogurt produced in week i, as wellas any
yogurt already in storage, can be used to meet Yucky'sdemand 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_iand Y_i.

 

Output

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

 

Sample Input

4 5

88 200

89 400

97 300

91 500

 

Sample Output

126900

题目大意：

输入数据包含每天生产酸奶的价格和需要的数量，当天的需求可以当天生产，也可以用之前剩下的，每保存一天酸奶，每单位话费 样例中的 5。求最小花费。

思路：

每次生产时比较当天的价格和之前的价格（加上保存到现在的花费），取小的就行啦。

感想：

大半夜写的，困死了。。

AC代码：

#include

#include

#include

#include

#include

#include

#include

#include

#include

#include

#include

#include

using namespace std;

int a[10005];

int y[10005];

int main()

{

   //freopen("r.txt", "r", stdin);

    long longsum=0;

    intday,s,i;

   cin>>day>>s;

   for(i=1;i<=day;i++)

    {

       cin>>a[i];

       cin>>y[i];

       if(i==1)

       {

           sum+=a[i]*y[i];

           continue;

       }

       if(a[i-1]+s

           sum+=(a[i-1]+s)*y[i];

       else

       {

           sum+=a[i]*y[i];

       }

    }

   cout<<sum<<endl;

}