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.
牛们收购了一个奶酪工厂,接下来的N个星期里，牛奶价格和劳力价格不断起伏．第i周，生产一个单位奶酪需要Ci(1≤Ci≤5000)便士．工厂有一个货栈，保存一单位奶酪，每周需要S(1≤S≤100)便士，这个费用不会变化．货栈十分强大，可以存无限量的奶酪，而且保证它们不变质．工厂接到订单，在第i周需要交付Yi(0≤Yi≤104)单位的奶酪给委托人．第i周刚生产的奶酪，以及之前的存货，都可以作为产品交付．请帮牛们计算这段时间里完成任务的最小代价．
* 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.
第1行输入两个整数N和S．接下来N行输入Ci和Yi．
* 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.
输出最少的代价．注意，可能超过32位长整型．
Gold
题解：很经典的递推题（根本算不上DP，甚至有几分贪心的味道在里面），第五次见到了，不多说了，直接上代码（HansBug：难以想象这种题居然上usaco金组= =）
1 /**************************************************************
2 Problem: 1740
3 User: HansBug
4 Language: Pascal
5 Result: Accepted
6 Time:28 ms
7 Memory:224 kb
8 ****************************************************************/
9
10 var
11 i,j,k,l,m,n:longint;
12 ans:int64;
13 function min(x,y:longint):longint;
14 begin
15 if x<y then min:=x else min:=y;
16 end;
17 begin
18 readln(n,m);
19 l:=maxlongint-m;
20 for i:=1 to n do
21 begin
22 readln(j,k);
23 l:=min(l+m,j);
24 ans:=ans+int64(l)*int64(k);
25 end;
26 writeln(ans);
27 readln;
28 end.
