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NYOJ 215 Sum
Sum
时间限制:1000 ms | 内存限制:65535 KB
难度: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.- 输入
- 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. - 输出
- The output will contain the minimum number N for which the sum S can be obtained.
- 样例输入
3 12 0
- 样例输出
2 7
AC码:
#include<stdio.h> 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; }
NYOJ 215 Sum
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