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 a

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.

The first line of input contains integer n

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.

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

3
1 3
2 2
3 1

10

3
1 3
2 1
3 2

8

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.

題解：不斷更新肉價的最小值

 1 #include <iostream>
 2 #include <algorithm>
 3 #include <cstring>
 4 #include <cstdio>
 5 #include <vector>
 6 #include <cstdlib>
 7 #include <iomanip>
 8 #include <cmath>
 9 #include <ctime>
10 #include <map>
11 #include <set>
12 using namespace std;
13 #define lowbit(x) (x&(-x))
14 #define max(x,y) (x>y?x:y)
15 #define min(x,y) (x<y?x:y)
16 #define MAX 100000000000000000
17 #define MOD 1000000007
18 #define pi acos(-1.0)
19 #define ei exp(1)
20 #define PI 3.141592653589793238462
21 #define INF 0x3f3f3f3f3f
22 #define mem(a) (memset(a,0,sizeof(a)))
23 typedef long long ll;
24 const int N=100005;
25 const int mod=1e9+7;
26 int a[N],b[N];
27 int main()
28 {
29     int n;
30     scanf("%d",&n);
31     for(int i=0;i<n;i++){
32         scanf("%d %d",&a[i],&b[i]);
33     }
34     ll s=0;
35     s+=a[0]*b[0];
36     int t=b[0];
37     for(int i=1;i<n;i++){
38         if(b[i]<t) t=b[i];
39         s+=a[i]*t;
40     }
41     printf("%lld\n",s);
42     return 0;
43 }

Codeforce 588A - Duff and Meat （貪心）