zoukankan      html  css  js  c++  java
  • codeforces #326 div 2 A. Duff and Meat(水)

    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 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 forn 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.

    做点水题换换脑子。。。

    浙大月赛题已经做蒙了QAQ

    这题很简单。。

    扫的时候记得更新价钱的最小值就好了。

     1 /*************************************************************************
     2     > File Name: code/cf/#326/A.cpp
     3     > Author: 111qqz
     4     > Email: rkz2013@126.com 
     5     > Created Time: 2015年10月22日 星期四 20时00分33秒
     6  ************************************************************************/
     7 
     8 #include<iostream>
     9 #include<iomanip>
    10 #include<cstdio>
    11 #include<algorithm>
    12 #include<cmath>
    13 #include<cstring>
    14 #include<string>
    15 #include<map>
    16 #include<set>
    17 #include<queue>
    18 #include<vector>
    19 #include<stack>
    20 #include<cctype>
    21                  
    22 #define yn hez111qqz
    23 #define j1 cute111qqz
    24 #define ms(a,x) memset(a,x,sizeof(a))
    25 using namespace std;
    26 const int dx4[4]={1,0,0,-1};
    27 const int dy4[4]={0,-1,1,0};
    28 typedef long long LL;
    29 typedef double DB;
    30 const int inf = 0x3f3f3f3f;
    31 const int N=1E5+7;
    32 int n;
    33 int a[N],p[N];
    34 int main()
    35 {
    36   #ifndef  ONLINE_JUDGE 
    37    freopen("in.txt","r",stdin);
    38   #endif
    39 
    40    scanf("%d",&n);
    41    for ( int i = 0 ;  i < n ; i++) scanf("%d %d",&a[i],&p[i]);
    42    int mi = inf;
    43    int ans = 0 ;
    44    for ( int i = 0 ; i < n ; i++)
    45    {
    46        if (p[i]<mi)
    47        mi = p[i];
    48        
    49        ans += mi*a[i];
    50    }
    51    cout<<ans<<endl;
    52   
    53    
    54  #ifndef ONLINE_JUDGE  
    55   fclose(stdin);
    56   #endif
    57     return 0;
    58 }
    View Code
  • 相关阅读:
    九度OJ 1035:找出直系亲属(二叉树)
    while 循环的理解
    while 循环的理解
    count 变量的使用
    count 变量的使用
    数据结构与算法的实现 —— 结点定义与数据结构的选择
    数据结构与算法的实现 —— 结点定义与数据结构的选择
    最小生成树(MST,minimum spanning tree)
    最小生成树(MST,minimum spanning tree)
    Opencv中Mat矩阵相乘——点乘、dot、mul运算详解
  • 原文地址:https://www.cnblogs.com/111qqz/p/4902678.html
Copyright © 2011-2022 走看看