ACM Yogurt factory(挑战程序设计竞赛)

Yogurt factory

Time Limit: 1000ms

Memory Limit: 65536KB

This problem will be judged on  PKU. Original ID:  2393
64-bit integer IO format:  %lld      Java class name:  Main

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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. 

Source

USACO 2005 March Gold

#include <iostream>
#include <cstdio>

using namespace std;

#define MAX_N 20000

int n,s;
long long c,y;

int main()
{
    scanf("%d%d",&n,&s);

    long long c,y;

    cin>>c>>y;
    long long last=c;
    long long ans=c*y;
    for(int i=1;i<n;i++)
    {
        cin>>c>>y;
        if(last+s<c)
        {
            ans+=(last+s)*y;
            last+=s;
        }
        else
        {
            ans+=c*y;
            last=c;
        }
    }

    cout<<ans<<endl;

    return 0;
}