标签：style blog color os io for ar art
题目描述：
新入职华为的小伙伴们都有在oj上面刷题的任务，共需要刷100道初级题，45道中级题，5道高级题,其中，做出来的高级题如果超标可以当初级或者中级题，做出来的中级题如果超标可以当初级题。每天，出题的大哥会给大家出Xi道题，这Xi道题属于同一个难度级别，小伙伴们要么用一天时间把这些题全做出来，要么就不做。现在，给你每天出题大哥出的题数以及难度，请问，小伙伴们最少要挑选其中几天去做题，才能把这150道题的任务完成呢？
输入示例：
5
100 70 5 5 55
1 2 2 2 3
输出：
2
表示两天就可以完成。
自己的解决思路：
#include <iostream>
#include <fstream>
#include <string>
#include <vector>
#include <algorithm>
using namespace std;
#define LOWLEVEL 100
#define MIDLEVEL 45
#define UPLEVEL 5
bool MoreThan(int a,int b)
{
return a > b;
}
int theMax(int a,int b,int c)
{
return (a>b?(a>c?a:c):(b>c?b:c));
}
void display(char* str,vector<int> &src)
{
int i,n;
n = src.size();
cout<<str<<" : ";
for (i = 0; i < n; i++)
{
cout<<src[i]<<" ";
}
cout<<endl;
}
void OJDoWorks(int days,vector<int> &numbers,vector<int> &nandu)
{
int i,n;
int dayOver,flagu,flagm,flagl;
int lowsum,midsum,upsum;
vector<int> lowv, midv, upv,sortv,sortlv;
n = numbers.size();
flagl = flagm = flagu = -1;
lowsum = midsum = upsum = dayOver = 0;
//将题目进行分类
for (i = 0; i < n; i++)
{
if (nandu[i] == 1)
{
lowv.push_back(numbers[i]);
}
else if (nandu[i] == 2)
{
midv.push_back(numbers[i]);
}
else
{
upv.push_back(numbers[i]);
}
}
//按大小进行排序
sort(upv.begin(),upv.end(),MoreThan);
sort(midv.begin(),midv.end(),MoreThan);
sort(lowv.begin(),lowv.end(),MoreThan);
display("up",upv);
display("mid",midv);
display("low",lowv);
n = theMax(upv.size(),midv.size(),lowv.size());
//先把高级的做满
for(i = 0; i < upv.size();i++)
{
dayOver++;
upsum += upv[i];
if (upsum >= UPLEVEL)
{
flagu = i;
break;
}
}
//高级题目不够
if (flagu < 0)
{
return;
}
midsum = upsum - UPLEVEL;
//将剩下的高级和中级的题目进行排序
for (i = flagu+1; i< upv.size(); i++)
{
sortv.push_back(upv[i]);
}
for (i = 0; i < midv.size(); i++)
{
sortv.push_back(midv[i]);
}
sort(sortv.begin(),sortv.end(),MoreThan);
display("sort mid:",sortv);
//再把中级的做满
n = sortv.size();
for (i = 0; i < n; i++)
{
if (midsum >= MIDLEVEL)
{
flagm = i;
break;
}
dayOver++;
midsum += sortv[i];
}
//中级题目数量不够
if (flagm < 0)
{
return;
}
lowsum = midsum - MIDLEVEL;
//再做低级的题目
for ( i = flagm; i < n; i++)
{
sortlv.push_back(sortv[i]);
}
for (i = 0; i < lowv.size(); i++)
{
sortlv.push_back(lowv[i]);
}
sort(sortlv.begin(),sortlv.end(),MoreThan);
display("sort low",sortlv);
n = sortlv.size();
for (i = 0; i < n; i++)
{
if (lowsum >= LOWLEVEL)
{
flagl = i;
break;
}
dayOver++;
lowsum += sortlv[i];
}
if ((upsum >= UPLEVEL) && (midsum >= MIDLEVEL) && (lowsum >= LOWLEVEL))
{
cout<<dayOver<<endl;
}
else
{
return;
}
}
int main()
{
//fstream in("data.txt");
int n,i,value;
vector<int> numbers,nandu;
cin>>n;
for (i = 0; i < n; i++)
{
cin>>value;
numbers.push_back(value);
}
for (i = 0; i < n; i++)
{
cin>>value;
nandu.push_back(value);
}
OJDoWorks(n,numbers,nandu);
cout<<endl;
return 0;
}
标签：style blog color os io for ar art
原文地址：http://www.cnblogs.com/xingma0910/p/3917979.html
