题目：
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 5 88 200 89 400 97 300 91 500
Sample Output
126900
翻译：
奶牛们有一个工厂用来生产奶酪，接下来的N周时间里，在第i周生产一份的奶酪需要花费Ci，同时它们也有一个储存室，奶酪放在那永远不会坏，并且可以无限放，每一单元奶酪放在那的价格恒定为每周s。然后奶牛在第i周会交付顾客Yi的奶酪，让你求最小花费。
思路：
只维护一个最小值，因为如果牛奶要保存，必须选取的是当前费用里，最小的哪一个。而它最小，就永远比之前的费用要小。
AC代码：
#include<iostream>
#include<algorithm>
using namespace std;
#pragma warning(disable:4996);
#define mem(sx,sy) memset(sx,sy,sizeof(sx))
#define INF 0x3f3f3f3f
typedef long long ll;
const ll MAXN = 1e6 + 5;
#define MAXSIZE 100
int main() {
int n, s;
while (cin >> n >> s) {
int pre = INF;
ll sum = 0;
for (int i = 0; i < n; i++) {
int c, y;
cin >> c >> y;
pre = min(pre, c);
sum += y * pre;
pre += s;
}
cout << sum << endl;
}
}