Challenges of CSDN programming-Gaussian Formula

Gaussian formula details:

Gauss invented the Sum Formula of the arithmetic difference series in elementary school: 1 + 2 +... + 100 = 5050. Now the question is to give you a positive integer n. How many continuous positive integers can you represent? (Itself ).

Input Format:

Multiple groups of data, one row of each group of data, a positive integer n. 0 <n <2000000000

Output Format:

A row of data in each group contains a positive integer, indicating the result.

Q & A description:

Input example

5

120

Output example:

2

4

Explanation:

5 = 2 + 3 = 5

120 = 1 + 2 +... + 15 = 22 + 23 + 24 + 25 + 26 = 39 + 40 + 41 = 120

Draft Code:

/* 3:3, 1 + 2 = 34: 4, 5:5, 2 + 3 = 56: 6, 1 + 2 + 3 = 6 7: 7, 3 + 4 =, 4 + 5 = 9, 2 + 3 + 4 = 910: 10, 1 + 2 + 3 + 4 = 10 analysis: input n to set s, x (x continuous numbers starting from s add n, for example: 10 = 1 + 2 + 3 + 4 n: 10 s: 1 x = 4) so there is a summation formula: (s + x-1) ----------- * x = n2 according to the summation formula brute force to solve */# include "stdio. h "# include" math. h "int main () {int count; long x, n; while (scanf (" % I64d ", & n )! = EOF) {count = 1; for (x = 2; x <n; x ++) {double s = (n + (x-x)/2.0) /x; if (s> = 1) {if (floor (s + 0.5) = s) {// judge s as an integer printf ("% d> x: % I64d s: % d \ n ", count + 1, x, (int) s); count ++;} else {printf (" % d, % lf not fully divided! \ N ", x, s) ;}} else {printf (" a maximum of % d is allowed! \ N ", x); break;} printf (" result: % d \ n ", count); break;} return 0 ;}

AC post code:

#include "stdio.h"#include "math.h"int main(){int count;long long x,n;while(scanf("%I64d",&n)!=EOF){double s=n;for(count=0,x=2;s>=1;x++){if(floor(s+0.5)==s){count++;//printf("%d >  x:%d  s:%d\n",count,x-1,(int)s);}s=(double)n/x+(1-x)/2.0;}printf("%d\n",count);//break;} return 0;}

Note: When the data type in the program is int, it may overflow during the calculation of relatively large data.

CSDN challenge programming exchange group: 372863405