H - Yogurt factory

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

Inputcopy	Outputcopy
4 5 88 200 89 400 97 300 91 500	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.

题目描述

小 T 开办了一家机器工厂，在 N个星期内，原材料成本和劳动力价格不断起伏，第 i 周生产一台机器需要花费 Ci​ 元。若没把机器卖出去，每保养一台机器，每周需要花费 SS 元，这个费用不会发生变化。

机器工厂接到订单，在第 ii 周需要交付 Yi​ 台机器给委托人，第 i 周刚生产的机器，或者之前的存货，都可以进行交付。

请你计算出这 n 周时间内完成订单的最小代价。

输入格式

第一行输入两个整数 N 和 S，接下来 N 行每行两个数 Ci​ 和 Yi​。

输出格式

输出一个整数，表示最少的代价。

#include
#include
using namespace std;
typedef long long ll;

int main()
{
	int n, s; cin >> n >> s;
	ll last=0, sum=0;
	for (int i = 1; i <= n; i++)
	{
		ll c, y; scanf_s("%lld%lld", &c, &y);
		if (i == 1) last = c;
		else last = min(last + s, c);
		sum += last * y;
	}
	cout << sum << endl;
}