Yogurt factory

Yogurt factory

Crawling in process...Crawling failedTime Limit:1000MSMemory Limit:65536KB 64bit IO Format:%I64d & %I64u

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

4 588 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. 

题目大意：由于市场原因牛奶的生产价格会有所波动，要求在将来的n周里，该公司拥有一个足够大的仓库储存制好的牛奶，每个单位的牛奶每周储存

 所需的费用为s，现给出将来n周生产每单位牛奶的价格及客户所需的数量，求出该公司将来n周内所要支付的最小生产值

题解:由于每周的生产价格波动，仓库的价格不变，所以只须拿当前周的生产价格与上周的比....唯一要注意的就是由于数据较大要用—int64类型

#include<cstdio>#define min(a,b) (a>b?b:a)const int M=10010;int c[M],y[M];int n,s,i,j;__int64 res;using namespace std;int main(){    c[0]=999999;    while(scanf("%d%d",&n,&s)!=EOF)    {        for(i=1;i<=n;i++)        {            scanf("%d%d",&c[i],&y[i]);            c[i]=min(c[i],c[i-1]+s);        }        res=0;        for(i=1;i<=n;i++)         res+=c[i]*y[i];        printf("%I64d\n",res);    }    return 0;}

                  

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