【Codeforces Round 326 (Div 2)A】【贪心】Duff and Meat 屯肉前溯花费最低

A. Duff and Meat

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Duff is addicted to meat! Malek wants to keep her happy for n days. In order to be happy in i-th day, she needs to eat exactly a_ikilograms of meat.

There is a big shop uptown and Malek wants to buy meat for her from there. In i-th day, they sell meat for p_i dollars per kilogram. Malek knows all numbers a₁, ..., a_n and p₁, ..., p_n. In each day, he can buy arbitrary amount of meat, also he can keep some meat he has for the future.

Malek is a little tired from cooking meat, so he asked for your help. Help him to minimize the total money he spends to keep Duff happy for n days.

Input

The first line of input contains integer n (1 ≤ n ≤ 10⁵), the number of days.

In the next n lines, i-th line contains two integers a_i and p_i (1 ≤ a_i, p_i ≤ 100), the amount of meat Duff needs and the cost of meat in that day.

Output

Print the minimum money needed to keep Duff happy for n days, in one line.

Sample test(s)

input

3
1 3
2 2
3 1

output

10

input

3
1 3
2 1
3 2

output

8

Note

In the first sample case: An optimal way would be to buy 1 kg on the first day, 2 kg on the second day and 3 kg on the third day.

In the second sample case: An optimal way would be to buy 1 kg on the first day and 5 kg (needed meat for the second and third day) on the second day.

#include<stdio.h>
#include<string.h>
#include<ctype.h>
#include<math.h>
#include<iostream>
#include<string>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre(){freopen("c://test//input.in","r",stdin);freopen("c://test//output.out","w",stdout);}
#define MS(x,y) memset(x,y,sizeof(x))
#define MC(x,y) memcpy(x,y,sizeof(x))
#define MP(x,y) make_pair(x,y)
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1,class T2>inline void gmax(T1 &a,T2 b){if(b>a)a=b;}
template <class T1,class T2>inline void gmin(T1 &a,T2 b){if(b<a)a=b;}
const int N=0,M=0,Z=1e9+7,ms63=1061109567;
int casenum,casei;
int main()
{
	int n;
	while(~scanf("%d",&n))
	{
		int P=1e9;
		int ans=0;
		for(int i=1;i<=n;i++)
		{
			int num,pri;
			scanf("%d%d",&num,&pri);
			gmin(P,pri);
			ans+=num*P;
		}
		printf("%d\n",ans);
	}
	return 0;
}
/*
【题意】
有n（1e5）天时间，
对于第i天，我们要消耗num[i]单位的食物，
这天如果选择买食物，1单位食物的单价为pri[i]，
每天买的食物数量都没有上限，可以囤积到最后。
问你满足这n天的食物需求，最少需要花多少钱。

【类型】
贪心

【分析】
首先，我们发现，如果某天买食物的单价比之后的都低，那我们肯定以这个价买。
这样从前向后有一个阶段性的延展，思考起来和实现起来都有些困难。

于是，我们可以变化一下，做从后向前的思维。
对于每天食用的食物，价格肯定是在它之前的最低价买的。
于是更新一个前缀最低价，这道题就做完了。

*/