网易笔试题:输入一个正整数,若该数能用几个连续正整数之和表示,则输出所有可能的正整数序列。 - 飞鸟快快的专栏 - 博客频道 - CSDN.NET


网易笔试题:输入一个正整数,若该数能用几个连续正整数之和表示,则输出所有可能的正整数序列。 - 飞鸟快快的专栏 - 博客频道 - CSDN.NET

编程题:输入一个正整数,若该数能用几个连续正整数之和表示,则输出所有可能的正整数序列。

解题思路:
          **找到数学规律。n以i为开头的j个整数的和为i*j+j*(j-1)/2;判断输入的整数是否和计算结果相同。
          如果相同则循环输出以i为开头的j个整数。
          这个题目给我最大的启示就是要学会用数学的方法来解决问题,在此之前我一直希望通过一种递归或者循环
          来建立一个包含所有能被连续整数和表示的整数集合,这我个人认为是一种计算机式的思维。
          以后在解决问题的时候要尝试数学抽象来解决问题。
public  static   void  print(int  n){
int  r = 0;
for(int i = 1; i< (n+1)/2;  i++){
for(int  j = 1; j< (n+1)/2; j++){
r = i*j + j*(j-1)/2;
if(r == n){
System.out.println("存在从  " +i +"  开始的连续   "+j+"个数字,使得它们的和为  "+ n);
return;
}
}//inner for
}//  outer  for

System.out.println("不存在连续的数字使得它们的和为   "+ n);
}
http://www.5uzh.com/view.aspx?kn=4007878
http://bylijinnan.iteye.com/blog/1399259
一个数M若可以写成以a开头的连续n个自然数之和,则M=a+(a+1)+(a+2)+…+(a+n-1)=n*a+n*(n-1)/2,要求a!=0,否则就是以a+1开头的连续n-1个整数了,也就是要求(M-n*(n-1)/2)%n==0,这样就很容易判断一个数可不可以写成连续n个自然数的形式了,遍历n=2…sqrt(M)*2,还可以输出所有解。
void divide(int num)  
{  
    int i,j,a;  
    for(i=2; i<=sqrt((float)num)*2; ++i)  
    {  
        if((num-i*(i-1)/2)%i==0)  
        {  
            a=(num-i*(i-1)/2)/i;  
            if(a>0)  
            {  
                for(j=0; j<i; ++j)  
                    cout<<a+j<<" ";  
            }  
            cout<<endl;  
        }  
    }   
}  
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