[POJ]1844 Sum

问题

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

12

Sample Output

7

分析

纯数学推导。。。。涉及到小学时学的高斯求和问题，即

sum(i)=i∗(i+1)2

sum(i) 是 i 所能确定的最大的数。题目肯定需要先满足：sum(i)>S，不然必然无解。
因此情况就分为：
1. sum(i)=S，此情况即为解
2. sum(i)>S，只需找到满足（sum(i)−S) 的 i 值，即可。其实情况1为情况2的特殊情况。对于一个确定的 i 值，所能确定的所有和的差值必然为2的倍数 （sum(i)−2∗x)。

因此，对于输入的S，可以先确定使得 sum(i)≥S 成立的最小 i。若 （sum(i)−S) 则 i 为解。否则，若 i 为奇数，则为了保证奇偶性，需要 i+2

sum(i)−S=i∗(i+1)2−S为奇数

(i+2）∗(i+3)2−S=i∗(i+1)2−S+2∗i+3为偶数

若 i 为偶数，则为了保证奇偶性，需要 i+1

sum(i)−S=i∗(i+1)2−S为奇数

(i+1）∗(i+2)2−S=i∗(i+1)2−S+i+1为偶数

源代码

#include <iostream>#include <cmath>using namespace std;int main() {    int S, i;    while (cin >> S) {        i = (int)sqrt(2.0 * S);        if (i * (i + 1) < 2 * S)            ++i;        if ((i * (i + 1) / 2 - S) % 2 == 0)            cout << i << endl;        else {            if (i % 2 == 1)                cout << i + 2 << endl;            else                cout << i + 1 << endl;        }    }    return 0;}

程序结果

Result Memory Time Language Code Length Accepted 256K 0MS C++ 333B