C

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

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.

这应该算贪心的思想吧，刚做不清楚；
就是从这周开始，要不要把接下来几天的也生产掉，对比一下花费就可以了

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<cmath>
#include<string.h>
#include<algorithm>
#define sf scanf
#define pf printf
#define pb push_back
#define mm(a,b) memset((a),(b),sizeof(a))
#include<vector>
const double pi=acos(-1.0);
typedef __int64 ll;
typedef long double ld;
const ll MOD=1e9+7;
using namespace std;
int num[10005],n,s,cost[10005],d[10005];
int main()
{
	mm(num,0);
	mm(cost,0);
	mm(d,0);
	cin>>n>>s;
	for(int i=0;i<n;i++)
	sf("%d%d",&cost[i],&num[i]);
	for(int i=0;i<n;i++)
	{
		int ans=0;
		for(int j=1;j<n-i;j++)
		{
			if(cost[i]*num[i+j]+num[i+j]*j*s<cost[i+j]*num[i+j])
			{
				ans++;
			}else
			break;
		}
		if(ans)
		{
			
			for(int j=0;j<=ans;j++)
			{
				d[i]+=num[i+j];
			}
			i+=ans;
		}else
		d[i]=num[i];
	}
	ll sum=0,tot=0;
	for(int i=0;i<n;i++)
	{
		sum+=d[i]*cost[i]+tot*s;
		tot=tot+d[i]-num[i];
		}
	
	pf("%I64d",sum);
	return 0;
}