Educational Codeforces Round 3 D. Gadgets for dollars and pounds 二分+前缀
2 seconds
256 megabytes
standard input
standard output
Nura wants to buy k gadgets. She has only s burles for that. She can buy each gadget for dollars or for pounds. So each gadget is selling only for some type of currency. The type of currency and the cost in that currency are not changing.
Nura can buy gadgets for n days. For each day you know the exchange rates of dollar and pound, so you know the cost of conversion burles to dollars or to pounds.
Each day (from 1 to n) Nura can buy some gadgets by current exchange rate. Each day she can buy any gadgets she wants, but each gadget can be bought no more than once during n days.
Help Nura to find the minimum day index when she will have k gadgets. Nura always pays with burles, which are converted according to the exchange rate of the purchase day. Nura can't buy dollars or pounds, she always stores only burles. Gadgets are numbered with integers from 1 to m in order of their appearing in input.
First line contains four integers n, m, k, s (1 ≤ n ≤ 2·105, 1 ≤ k ≤ m ≤ 2·105, 1 ≤ s ≤ 109) — number of days, total number and required number of gadgets, number of burles Nura has.
Second line contains n integers ai (1 ≤ ai ≤ 106) — the cost of one dollar in burles on i-th day.
Third line contains n integers bi (1 ≤ bi ≤ 106) — the cost of one pound in burles on i-th day.
Each of the next m lines contains two integers ti, ci (1 ≤ ti ≤ 2, 1 ≤ ci ≤ 106) — type of the gadget and it's cost. For the gadgets of the first type cost is specified in dollars. For the gadgets of the second type cost is specified in pounds.
If Nura can't buy k gadgets print the only line with the number -1.
Otherwise the first line should contain integer d — the minimum day index, when Nura will have k gadgets. On each of the next k lines print two integers qi, di — the number of gadget and the day gadget should be bought. All values qi should be different, but the values di can coincide (so Nura can buy several gadgets at one day). The days are numbered from 1 to n.
In case there are multiple possible solutions, print any of them.
5 4 2 2
1 2 3 2 1
3 2 1 2 3
1 1
2 1
1 2
2 2
3
1 1
2 3
4 3 2 200
69 70 71 72
104 105 106 107
1 1
2 2
1 2
-1
4 3 1 1000000000
900000 910000 940000 990000
990000 999000 999900 999990
1 87654
2 76543
1 65432
-1
题意:你有n个天数的汇率,现在你有s元人民币,第i天可以随意换a[i]个人民币换1美元,b[i]个人民币换1英镑,但是不可以换了存起来,必须今天用掉;
下面有m个东西,q表示只能用美元或者英镑买,需要的花费,你需要求最小的天数使得你买任意k个;
思路:前缀最小值,存两个汇率最小的时候,分别在这两天买;
将美元可以买跟英镑可以买的分开。。算花费最少的人民币可以买到东西;
二分天数即可;
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-14
#define bug(x) cout<<"bug"<<x<<endl;
const int N=2e5+,M=1e6+,inf=1e9+;
const ll INF=1e18+,mod=; ///数组大小
vector<pair<int,int> >v[];
pair<int,int> mina[N],minb[N];
int check(int x,int k,int s)
{
int a=mina[x].first;
int b=minb[x].first;
int sa=,sb=,si=;
while()
{
if(sa==v[].size()&&sb==v[].size())break;
if(sa==v[].size())
{
if(s<1LL*v[][sb].first*b)
break;
else
s-=v[][sb].first*b,sb++,si++;
}
else if(sb==v[].size())
{
if(s<1LL*v[][sa].first*a)
break;
else
s-=v[][sa].first*a,sa++,si++;
}
else
{
if(1LL*v[][sa].first*a<=1LL*v[][sb].first*b)
{
if(s<1LL*v[][sa].first*a)
break;
else
s-=v[][sa].first*a,sa++,si++;
}
else
{
if(s<1LL*v[][sb].first*b)
break;
else
s-=v[][sb].first*b,sb++,si++;
}
}
}
if(si>=k)return ;
return ;
}
void output(int x,int k,int s)
{
int a=mina[x].first;
int b=minb[x].first;
int sa=,sb=,si=;
while(si<k)
{
if(sa==v[].size())
{
if(s<1LL*v[][sb].first*b)
break;
else
s-=v[][sb].first*b,printf("%d %d\n",v[][sb].second,minb[x].second),sb++,si++;
}
else if(sb==v[].size())
{
if(s<1LL*v[][sa].first*a)
break;
else
s-=v[][sa].first*a,printf("%d %d\n",v[][sa].second,mina[x].second),sa++,si++;
}
else
{
if(1LL*v[][sa].first*a<=1LL*v[][sb].first*b)
{
if(s<1LL*v[][sa].first*a)
break;
else
s-=v[][sa].first*a,printf("%d %d\n",v[][sa].second,mina[x].second),sa++,si++;
}
else
{
if(s<1LL*v[][sb].first*b)
break;
else
s-=v[][sb].first*b,printf("%d %d\n",v[][sb].second,minb[x].second),sb++,si++;
}
}
}
}
int main()
{
int n,k,m,s;
scanf("%d%d%d%d",&n,&m,&k,&s);
mina[]=make_pair(inf,);
minb[]=make_pair(inf,);
for(int i=;i<=n;i++)
{
int x;
scanf("%d",&x);
if(x<mina[i-].first)
mina[i]=make_pair(x,i);
else mina[i]=mina[i-]; }
for(int i=;i<=n;i++)
{
int y;
scanf("%d",&y);
if(y<minb[i-].first)
minb[i]=make_pair(y,i);
else minb[i]=minb[i-];
}
for(int i=;i<=m;i++)
{
int t,x;
scanf("%d%d",&t,&x);
v[t].push_back(make_pair(x,i));
}
for(int i=;i<=;i++)
sort(v[i].begin(),v[i].end());
int st=,en=n,ans=-;
while(st<=en)
{
int mid=(st+en)>>;
if(check(mid,k,s))
{
ans=mid;
en=mid-;
}
else
st=mid+;
}
printf("%d\n",ans);
if(ans!=-)
{
output(ans,k,s);
}
return ;
}
Educational Codeforces Round 3 D. Gadgets for dollars and pounds 二分+前缀的更多相关文章
- Codeforces Educational Codeforces Round 3 D. Gadgets for dollars and pounds 二分,贪心
D. Gadgets for dollars and pounds 题目连接: http://www.codeforces.com/contest/609/problem/C Description ...
- CF# Educational Codeforces Round 3 D. Gadgets for dollars and pounds
D. Gadgets for dollars and pounds time limit per test 2 seconds memory limit per test 256 megabytes ...
- CodeForce---Educational Codeforces Round 3 D. Gadgets for dollars and pounds 正题
对于这题笔者无解,只有手抄一份正解过来了: 基本思想就是 : 二分答案,对于第x天,计算它最少的花费f(x),<=s就是可行的,这是一个单调的函数,所以可以二分. 对于f(x)的计算,我用了nl ...
- codeforces 609D D. Gadgets for dollars and pounds(二分+贪心)
题目链接: D. Gadgets for dollars and pounds time limit per test 2 seconds memory limit per test 256 mega ...
- Educational Codeforces Round 6 D. Professor GukiZ and Two Arrays 二分
D. Professor GukiZ and Two Arrays 题目连接: http://www.codeforces.com/contest/620/problem/D Description ...
- Educational Codeforces Round 64 (Rated for Div. 2) (线段树二分)
题目:http://codeforces.com/contest/1156/problem/E 题意:给你1-n n个数,然后求有多少个区间[l,r] 满足 a[l]+a[r]=max([l, ...
- Educational Codeforces Round 61 (Rated for Div. 2)D(二分,模拟,思维)
#include<bits/stdc++.h>using namespace std;typedef long long ll;int n,k;ll a[200007],b[200007] ...
- Educational Codeforces Round 67 (Rated for Div. 2) B题【前缀+二分】【补题ING系列】
题意:给出一个字符串s, 可以从左往右拿走s的字符, 至少要到s的第几个位置才能拼成t 思路:用二维数组记录前缀,然后二分即可. #include<bits/stdc++.h> using ...
- Educational Codeforces Round 80 (Rated for Div. 2)D(二分答案,状压检验)
这题1<<M为255,可以logN二分答案后,N*M扫一遍表把N行数据转化为一个小于等于255的数字,再255^2检验答案(比扫一遍表复杂度低),复杂度约为N*M*logN #define ...
随机推荐
- GUI常用对象介绍3
%text hf = axes; ht = text(,,'示例'); get(ht); %公式 并且设置位置坐标 (积分符号) text('String','\int_0^x dF(x)','Pos ...
- C#使用SmtpClient发送邮件
目的:写一个可发送邮件的DLL. 原理: 例如A使用163邮箱发送邮件给B(qq邮箱).首先A会把邮件通过SMTP(Simple Mail Transfer Protocol)协议传输到163的Smt ...
- SQL注入(dvwa环境)
首先登录DVWA主页: 1.修改安全级别为LOW级(第一次玩别打脸),如图中DVWA Security页面中. 2.进入SQL Injection页面,出错了.(心里想着这DVWA是官网下的不至于玩不 ...
- 怎样从外网访问内网Node.js?
本地安装了一个Node.js,只能在局域网内访问,怎样从外网也能访问到本地的Node.js呢?本文将介绍具体的实现步骤. 1. 准备工作 1.1 安装并启动Node.js 默认安装的Node.js端口 ...
- Linux内核启动流程与模块机制
本文旨在简单的介绍一下Linux的启动流程与模块机制: Linux启动的C入口位于/Linux.2.6.22.6/init/main.c::start_kernel() 下图简要的描述了一下内核初始化 ...
- 基于 SSL 的 Nginx 反向代理
基于 SSL 的 Nginx 反向代理 描述: 线上zabbix因机房网络问题,外网接口无法对外访问,因此采用同机房的另外一台服务器做反向代理. 线上用于zabbix提供web访问的Nginx,采用h ...
- select2 AJAX获取数据
页面效果: index.html <!DOCTYPE html> <html> <head> <meta charset="utf-8"& ...
- Python3 解析excel文件
Python3 解析读取excel文件 一.第三方库 import xlrd 二.代码示例 import xlrd ''' 读取Excel每个sheet的第一列和第二列的值,拼接成json串,写入文件 ...
- js获取对象的key
var obj = {"name":"名字","age":"18"};var temp = "";f ...
- Windows Media Player添加播放插件