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

4 588 20089 40097 30091 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 20000int 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;}