51Nod 1090 3个数和为0 + Atcoder Tenka1 Programmer Contest N/4 暴力枚举+二分

思路都是一样的，枚举要求的结果中的一个数，其余的在此基础上二分，很常见的方法，也很容易想到，然而……还是要练啊，，，

大概……一般……也许……不会T吧……

51Nod 1090  3个数和为0 传送门

直接贴代码~

#include<iostream>   
#include<algorithm>
#include<vector> 
#include<string.h>
#include<stack>
using namespace std;
typedef long long ll;
const int MAX = 1e3 + 5;
int n;
ll a[MAX];
int main()
{
    ios::sync_with_stdio(false);
    while (cin >> n)
    {
        for (int i = 0; i<n; i++)
            cin >> a[i];
        int flag = 1;
        sort(a, a + n);
        for (int i = 0; i<n; i++) //暴力枚举a[i],a[j],二分a[k] 
        {
            for (int j = i + 1; j<n; j++)
            {
                ll x = -a[i] - a[j]; //记得用long long 不然会WA 
                int l = j + 1, r = n - 1, mid;
                while (l <= r)
                {
                    mid = (l + r) / 2;
                    if (a[mid] < x) l = mid+1;
                    else if(a[mid]>x) r = mid - 1;
                    else {
                        flag = 0;
                        cout << a[i] << " " << a[j] << " " << a[mid] << endl;
                        break;
                    }
                }
            }
        }
        if (flag)
            cout << "No Solution" << endl;
    }
    return 0;
}

//抄来的代码~感觉这个方法更好一些，以后如果有类似的三个数的问题可以用这个
#include<iostream>   
#include<algorithm>
#include<vector> 
#include<string.h>
#include<stack>
using namespace std;
typedef long long ll;
const int MAX=1e3+5;
int n,a[MAX];
int main()
{
    ios::sync_with_stdio(false );
    while(cin>>n)
    {
     for(int i=0;i<n;i++)
     cin>>a[i];
     int flag=1;
     sort(a,a+n); 
     for(int i=0;i<n;i++)
     {
         int l=i+1,r=n-1,x;
         while(l<r)
         {
           x=a[i]+a[l]+a[r];
           if(x>0) r--;
           else if(x<0) l++;
           else 
           {
               flag=0;
               cout<<a[i]<<" "<<a[l]<<" "<<a[r]<<endl;
               l++;r--;
          }    
        }
     }
     if(flag)
     cout<<"No Solution"<<endl;
    } 
    return 0;
}

Atcoder Tenka1 Programmer Contest C：4/N 传送门 

#include<iostream>   
#include<algorithm>
#include<vector> 
#include<string.h>
#include<stack>
using namespace std;
typedef long long ll;
const int MAX = 1e3 + 5;
ll n;
int main()
{
    ios::sync_with_stdio(false);
    while (cin >> n)
    {
        int flag = 0;
        for (ll h = 1; h < 3505; h++)
        {
            for (ll k = 1 ; k < 3505; k++)
            {
                if ((4 * h*k - n*k - n*h) != 0) //这里要注意啊……分母不能为0
                {
                    ll  w = (n*h*k) / (4 * h*k - n*k - n*h);
                    if (w > 0 && (n*h*k) % (4 * h*k - n*k - n*h) == 0)//这里WA了好多次……不太懂为什么正数和正数作除法会得负数……orz……一开始没判正负一直在WA
                    {
                        flag = 1;
                        cout << h << " " << k << " " << w << endl;
                        break;
                    }
                }
            }
            if (flag) break;
        }
    }
    return 0;
}