一天一道算法题--6.9--数学题

 1 // give a sum and n
 2 // so the arr is from 1 to n
 3 // sum = n*(a1+an)/2 = na1 + n*(n-1) / 2;
 4 // so a1 = ( sum - n(n-1)/2 )/n
 5 
 6 #include <iostream>
 7 using namespace std;
 8 
 9 void slove( int sum , int n )
10 {
11     int i , value , cnt , first;
12     if( sum<=0 || n<=0 )
13         return;
14     cnt = 0;
15     i = 1;
16     value = sum - i*(i-1)/2;
17     while( value>=i ) // because a1 = ( sum - n*(n-1)/2 ) /n  if this can be worked the value must >= n and because the element must >=1 you know the arr is from 1 to n
18     {
19         if( value%i==0 ) // the element is integer
20         {
21             first = value/i;
22             printf( "the %d arr:  ",cnt+1 );
23             for( int j = first ; j<first+i ; j++ )
24             {
25                 printf( "%d\t",j );
26             }
27             printf( "\n" );
28         }
29         i++;
30         cnt++;
31         value = sum - i*(i-1)/2;
32     }
33     printf( "cnt: %d\n",cnt );
34 }
35 
36 int main()
37 {
38     int sum , n;
39     while( ~scanf("%d %d",&sum,&n) )
40     {
41         slove( sum , n );
42         getchar();
43     }
44     return 0;
45 }