博客
关于我
强烈建议你试试无所不能的chatGPT,快点击我
The 2013 ACM-ICPC Asia Changsha Regional Contest - A
阅读量:5760 次
发布时间:2019-06-18

本文共 2372 字,大约阅读时间需要 7 分钟。

Alice's Print Service

Time Limit: 2 Seconds      
Memory Limit: 65536 KB

Alice is providing print service, while the pricing doesn't seem to be reasonable, so people using her print service found some tricks to save money.

For example, the price when printing less than 100 pages is 20 cents per page, but when printing not less than 100 pages, you just need to pay only 10 cents per page. It's easy to figure out that if you want to print 99 pages, the best choice is to print an extra blank page so that the money you need to pay is 100 × 10 cents instead of 99 × 20 cents.

Now given the description of pricing strategy and some queries, your task is to figure out the best ways to complete those queries in order to save money.

Input

The first line contains an integer T (≈ 10) which is the number of test cases. Then T cases follow.

Each case contains 3 lines. The first line contains two integers n, m (0 < n, m ≤ 105). The second line contains 2n integers s1, p1, s2, p2, ..., sn, pn (0=s1 < s2 < ... < sn ≤ 109, 109 ≥ p1 ≥ p2 ≥ ... ≥ pn ≥ 0). The price when printing no less than si but less than si+1 pages is pi cents per page (for i=1..n-1). The price when printing no less than sn pages is pn cents per page. The third line containing m integers q1 .. qm (0 ≤ qi ≤ 109) are the queries.

Output

For each query qi, you should output the minimum amount of money (in cents) to pay if you want to print qi pages, one output in one line.

Sample Input

12 30 20 100 100 99 100

Sample Output

010001000
#include 
#include
#include
#include
#include
#include
#include
#define Min(a,b) ((a)<(b)?(a):(b))using namespace std ;const int Max_N = 100008 ;typedef long long LL ;LL dp[Max_N] ;LL S[Max_N] ,P[Max_N] ;int N ,M ;int find_id(LL x){ int Left = 1 ; int Right = N ; int mid ; int ans = -1; while(Left<=Right){ mid = (Left + Right)>>1 ; if(S[mid] >= x){ Right = mid - 1 ; ans = mid ; } else Left = mid + 1 ; } return ans ;}int main(){ int T ; LL x; scanf("%d",&T) ; while(T--){ scanf("%d%d",&N,&M) ; for(int i = 1 ;i <= N ;i++) scanf("%lld%lld",&S[i],&P[i]) ; dp[N] = S[N] * P[N] ; for(int i = N-1 ;i >= 1 ;i--){ dp[i] = Min(S[i]*P[i],dp[i+1]) ; } while(M--){ scanf("%lld",&x) ; int id = find_id(x) ; // cout<
<<" "<
<

 

转载于:https://www.cnblogs.com/liyangtianmen/p/3440015.html

你可能感兴趣的文章
android4.0 的图库Gallery2代码分析(四) 之相册的数据处理以及显示
查看>>
事务——原子性、一致性、隔离性和持久性的理解
查看>>
MySQL内存调优
查看>>
測试文档和用户说明书
查看>>
POJ 2309 BST
查看>>
【extjs】 extjs5 Ext.grid.Panel 搜索示例
查看>>
Java并发编程:synchronized
查看>>
apache2添加模块和添加站点
查看>>
thinkphp分页
查看>>
内部排序-第10章-《数据结构题集》习题解析-严蔚敏吴伟民版
查看>>
Windows:常见问题
查看>>
Zend Optimizer not installed可能原因及解决方法
查看>>
在ubuntu/deepin/mint等系统中使用命令删除文件或文件夹
查看>>
Asp.Net Core 发布和部署(Linux + Jexus )
查看>>
【Linux】——ctags
查看>>
AtomicInteger类保证线程安全的用法
查看>>
中控考勤机WEB主动上报接收SERVER程序
查看>>
爹地,我找到了!,15个极好的Linux find命令示例
查看>>
[唐诗]1蝉-虞世南
查看>>
Microsoft SQL Server Query Processor Internals and Architecture
查看>>