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 程式師世界 >> 編程語言 >> C語言 >> C++ >> C++入門知識 >> NYOJ 215 Sum

NYOJ 215 Sum

編輯:C++入門知識

Sum

時間限制:1000 ms | 內存限制:65535 KB 難度:2
描述
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.
輸入
The input consists N test cases.
The only line of every test cases contains a positive integer S (0< S <= 100000) which represents the sum to be obtained.
A zero terminate the input.
The number of test cases is less than 100000.
輸出
The output will contain the minimum number N for which the sum S can be obtained.
樣例輸入
3
12
0
樣例輸出
2
7
AC碼:
#include
int main()
{
	int n,i,t;
	while(scanf("%d",&n)&&n)
	{
		i=1;
		t=1;
		while(t!=n)
		{
			i++;
			t=i*(i+1)/2;
			if(t>n)
			{
				if((t-n)%2==0)
					t=n;
			}
		}
		printf("%d\n",i);
	}
	return 0;
}


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