Alice's Print Service
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
1
2 3
0 20 100 10
0 99 100
Sample Output
0
1000
1000
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
using namespace std;
typedef long long LL; LL a[],b[];
LL f[]; void prepare(LL n)
{
LL Min=b[n]*a[n];
LL ans;
f[n]=Min;
for(LL i=n-;i>=;i--)
{
ans=a[i]*b[i];
if(Min>ans)
Min=ans;
f[i]=Min;
}
}
LL EF(LL x,LL l,LL r)
{
LL mid=(l+r)/;
while(l<r)
{
if(a[mid]>x)
r=mid-;
else if(a[mid]<x)
l=mid;
else if(a[mid]==x)
return mid;
mid=(l+r+)/;
}
return mid;
} int main()
{
LL T;
LL i,n,m,x,k;
scanf("%lld",&T);
while(T--)
{
scanf("%lld%lld",&n,&m);
for(i=;i<=n;i++)
{
scanf("%lld%lld",&a[i],&b[i]);
}
prepare(n);
while(m--)
{
scanf("%lld",&x);
k=EF(x,,n);
LL ans=x*b[k];
if(k+<=n && ans>f[k+]) ans=f[k+];
printf("%lld\n",ans);
}
}
return ;
}
最新文章
- Windows phone 8.1布局控件
- Linux压缩指令
- 第一个WP8程序,照相机
- javascript最新深度克隆对象方法
- [原博客] POJ 1740 A New Stone Game
- css3 2D变换 transform
- Hadoop集群出现no data node to stop的解决方案
- Multi-Objective Data Placement for Multi-Cloud Socially Aware Services---INFOCOM 2014
- [iOS Animation]-CALayer 图像IO
- 简单的视频采集demo
- 完美解决cannot resolve symbol servlet 的报错
- java既然存在gc线程,为什么还存在内存泄漏?
- htop的安装和使用
- jvm结构
- log4j 输出原始数据到指定日志文件
- SpringMVC 学习 九 SSM环境搭建 (二) Spring配置文件的编写
- 后台取IE的相关信息
- 理解Storm Metrics
- mariadb(mysql)从库relaylog损坏无法同步的处理方法
- dbms_random 包的使用