codeforces 181.div2 300A –Array 思維問題

題目連結

1. The product of all numbers in the first set is less than zero ( < 0).
2. The product of all numbers in the second set is greater than zero ( > 0).
3. The product of all numbers in the third set is equal to zero.
4. Each number from the initial array must occur in exactly one set.

意思就是從一大堆書中選取 3堆數，使得第一堆所有數乘機<0，第二堆數的成績>0，第三堆數的成績等於0

這道題困擾了我一個小時~~囧

我的演算法是 

先分成a,b,c 三堆，然後a存放大於0的，b存放小於0的，c存放0；

然後  （1）若a堆的個數=0，則從b堆中拿出兩個放到a堆；

           （2）  經過1操作後若b堆的個數為偶數，則取出一個放到c堆中

即可就是最後答案

#include<stdio.h>#include<string.h>#include<vector>#include<algorithm>#include<iostream>using namespace std;vector<int>first,second,third;int main(){    int n;    while(scanf("%d",&n)!=EOF){        first.clear();        second.clear();        third.clear();        int num;        for(int i=1;i<=n;i++){            scanf("%d",&num);                                      if (num<0){                second.push_back(num);             }            if(num>0){                first.push_back(num);            }            if(num==0){                third.push_back(num);            }        }                  if( first.size()==0 ){   //操作1             int sum=second.size()-1;             num=second[sum];             second.erase(second.end()-1);   //刪數應該是最後項-1；             first.push_back(num);             sum=second.size()-1;             num=second[sum];             second.erase(second.end()-1);                              first.push_back(num);                 }        int sum=second.size();        if( sum%2==0 ){     //操作2               num=second[sum-1];               second.erase(second.end()-1);                     third.push_back(num);        }                 sort(first.begin(),first.end());  //從小到大排序        sort(second.begin(),second.end());        sort(third.begin(),third.end());                printf("%d",second.size());        for(int i=0;i!=second.size();i++)            printf(" %d",second[i]);        printf("\n");                printf("%d",first.size());        for(int i=0;i!=first.size();i++)            printf(" %d",first[i]);        printf("\n");        printf("%d",third.size());        for(int i=0;i!=third.size();i++)            printf(" %d",third[i]);        printf("\n");    }    }