[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\n",ans); 21 return 0; 22 }