国产探花免费观看_亚洲丰满少妇自慰呻吟_97日韩有码在线_资源在线日韩欧美_一区二区精品毛片,辰东完美世界有声小说,欢乐颂第一季,yy玄幻小说排行榜完本

首頁 > 學院 > 開發設計 > 正文

588A. Duff and Meat

2019-11-14 09:16:08
字體:
來源:轉載
供稿:網友
A. Duff and Meattime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard 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.

Examplesinput
31 32 23 1output
10input
31 32 13 2output
8Note

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 <iostream>#define INF 1000000000using namespace std;int main(){    int n,a,p,ans=0,price=INF;    cin>>n;    for(int i=0;i<n;i++)    {        cin>>a>>p;        price=min(price,p);        ans+=price*a;    }    cout<<ans;    return 0;}


發表評論 共有條評論
用戶名: 密碼:
驗證碼: 匿名發表
主站蜘蛛池模板: 南安市| 台州市| 阿拉善盟| 威远县| 隆德县| 兴隆县| 玉屏| 呈贡县| 枣阳市| 喀喇| 堆龙德庆县| 福州市| 山阳县| 宁化县| 阿城市| 定南县| 延长县| 巴楚县| 布拖县| 高阳县| 和顺县| 杭锦旗| 仁布县| 苍山县| 乌恰县| 如皋市| 夹江县| 博野县| 建德市| 康马县| 庆城县| 大厂| 泸水县| 千阳县| 永顺县| 溆浦县| 独山县| 永宁县| 汝南县| 普兰县| 丹棱县|