1.链接地址:
http://bailian.openjudge.cn/practice/1844
http://poj.org/problem?id=1844
2.题目:
Sum
Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 10031 Accepted: 6564 Description
Consider the natural numbers from 1 to N. By associating to each number a sign (+ or -) and calculating the value of this expression we obtain a sum S. The problem is to determine for a given sum S the minimum number N for which we can obtain S by associating signs for all numbers between 1 to N.
For a given S, find out the minimum value N in order to obtain S according to the conditions of the problem.Input
The only line contains in the first line a positive integer S (0< S <= 100000) which represents the sum to be obtained.Output
The output will contain the minimum number N for which the sum S can be obtained.Sample Input
12Sample Output
7Hint
The sum 12 can be obtained from at least 7 terms in the following way: 12 = -1+2+3+4+5+6-7.Source
3.思路:
4.代码:
1 #include "stdio.h" 2 //#include "stdlib.h" 3 int main() 4 { 5 int m; 6 scanf("%d",&m); 7 int k = (int)((sqrt((float)(1)+8*m)+1)/2); 8 while((k*k+k-2*m)%4!=0) 9 { 10 k++; 11 } 12 printf("%d",k); 13 //system("pause"); 14 return 0; 15 }