【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

31 32 23 1

output

10

input

31 32 13 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|1typedef 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。*/