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Problem Description

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.< br>< br>* 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

4 588 20089 40097 30091 500

 

Sample Output

126900

 

这个题目有点类似老师讲的钓鱼的问题有点类似都是计算本月成本和上月的比较。当时看到数据量偏大存储就用了long long；

代码：

#include <iostream> 
using namespace std; 
struct node 
{
    int c,y; 
}week[10002]; 

int main() 
{ 
        int N,S; 
    while(cin >> N >> S) 
    { 
        for(int i = 0;i < N;i++) 
            cin >> week[i].c >> week[i].y; 
 
        long long cost = 0; 
     for(int i = 0;i < N;i++) 
       { 
           int temp = i,Y = 0; 
            int save = 0; 
            for(int j = temp;j < N && week[j].c - week[temp].c >= S*(j-temp);j++)             { 
               Y += week[j].y; 
               save += (Y-week[temp].y)*S; 
               i = j; 
           } 
            cost += Y*week[temp].c+save; 
       }         cout << cost << endl; 
   } 
   return 0;
   }