Codeforces Round #326 (Div. 2) A

Codeforces Round #326 (Div. 2) A http://codeforces.com/contest/588/problem/A

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 ai kilograms
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 pidollars
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

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.

简单分析，,Pi较下一个大的话，就买当天的，反之，Pi小的时候就把后面几天的都买下来，直到遇到下一个更小的pi。注意最后的结果超int范围了，用long long，但CF 的大数输出只能用%I64d。
有一个小技巧，Pi与之前的较小值m比较的时候，如果pi更小，则赋值给m，变量i减去1巧妙地在下一次循环中回到刚才的pi，m改变了，再一次比较。
#include<cstdio>

#include<cstring>

#include<iostream>

#include<algorithm>

#include<cmath>

#include<queue>

#include<cstdlib>

#include<string>

#define maxx 100010
using namespace std;

int a[maxx],p[maxx];

int main()

{

int n;

long long m,x;

while(~scanf("%d",&n))

{

x=0;

for(int i=1;i<=n;i++)

{

scanf("%d%d",&a[i],&p[i]);

}

m=p[1];

for(int i=1;i<=n; i++)

{

if(m<=p[i])

x+=a[i]*m;

else

{

m=p[i];

i-=1;

}

}

printf("%I64d\n",x);

}

return 0;

}