• [北大机试E]:Yogurt factory(模拟)


    总时间限制: 
    1000ms
     
    内存限制: 
    65536kB
    描述
    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.
    输入
    * 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.
    输出
    * 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.
    样例输入
    4 5
    88 200
    89 400
    97 300
    91 500
    样例输出
    126900
    提示
    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.

     模拟题,从最后倒着往前,如果相邻两个week的差大于S,说明在前一周多做然后store是赚的,否则就在这周再做。

     1 #include <cstdio>
     2 #include <iostream>
     3 #include <cmath>
     4 #include <cstdlib>
     5 using namespace std;
     6 int n,s;
     7 int c[10005],y[10005];
     8 int main(){
     9     scanf("%d%d",&n,&s);
    10     for (int i = 0;i < n;++i)
    11         scanf("%d%d",c+i,y+i);
    12     long long ans = 0;
    13     for (int i = n-1;i > 0;--i){
    14         if (c[i] - c[i-1] > s){//store
    15             ans += s*y[i];
    16             y[i-1] += y[i];
    17         }else ans += c[i]*y[i];
    18     } 
    19     ans += c[0]*y[0];
    20     printf("%lld
    ",ans);
    21     return 0;
    22 }
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  • 原文地址:https://www.cnblogs.com/mizersy/p/12233210.html
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