1015

题目编号：1015

题目简意：

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 5
88 200
89 400
97 300
91 500
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    126900
   
   
   
   

AC代码：

#include <iostream>

#include <algorithm>

 

using namespace std;

const int maxn = 10000 + 5;

 

int main()

{

    __int64 sum = 0;

    int n, s, c[maxn], y[maxn];

    cin >> n >> s;

    c[0] = maxn;

    for (int i = 1; i <= n; i++) {

        cin >> c[i] >> y[i];

        c[i] = min (c[i], c[i - 1] + s);

        sum += c[i] * y[i];

    }

    cout << sum << endl;

    return 0;

}