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Yogurt factory

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7798		Accepted: 3970

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.  

* 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<stdio.h>
int min(int x,int y)
{
    if(x<y) return x;
    return y;
}
int main()
{
    int n,s;
    __int64 ans;
    int C[10006],Y[10006];
    scanf("%d%d",&n,&s);
    C[0]=9999;
    for(int i=1;i<=n;++i)
    {
        scanf("%d%d",&C[i],&Y[i]);
        C[i]=min(C[i],C[i-1]+s);
    }
    ans=0;
    for(int i=1;i<=n;++i)
    {
            ans+=C[i]*Y[i];
    }
    printf("%I64d",ans);
    return 0;
}

/*
4 5
88 200
89 400