Duff and Meat（贪心）

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 aikilograms 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 pi dollars per kilogram. Malek knows all numbers a1, ..., an and p1, ..., pn. 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 ai and pi (1 ≤ ai, pi ≤ 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.

题意：

  就是求比它后而且比它单价高的替换。

思路：

  一开始想着从下到上去贪心，发现没有什么优化的，就这样子想了很久的优化，最后却发现如果从上到下直接就O(n)了，因为这样贪心是最优的。

AC代码：

#include<iostream>#include<algorithm>#include<cstring>#include<vector>#include<cstdio>#include<queue>using namespace std;#define CRL(a) memset(a,0,sizeof(a))typedef __int64 ll;#define T 100005struct node{    int food,m,num;}a[T];int main(){    /*freopen("input.txt","r",stdin);*/    int n,i,k,sum;    while(~scanf("%d",&n))    {        scanf("%d%d",&a[0].food,&a[0].m);k = a[0].m;        sum=a[0].food*a[0].m;        for(i=1;i<n;++i){            scanf("%d%d",&a[i].food,&a[i].m);            if(k<=a[i].m){a[i].m=k;sum+=k*a[i].food;}            else {k=a[i].m;sum+=k*a[i].food;}        }        printf("%d\n",sum);    }    return 0;}