[POJ]1844 Sum

[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

$$sum(i) = {i * ( i + 1 ) \over 2}$$
$sum(i)$ 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$（sum(i) - S ) % 2 == 0$ 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 $sum(i) \ge S$ 成立的最小 $i$。若 $（sum(i) - S ) % 2 == 0$ 则 $i$ 为解。否则，若 $i$ 为奇数，则为了保证奇偶性，需要 $i + 2$

$$sum(i) - S = {i * ( i + 1 ) \over 2} - S为奇数$$
$${(i + 2） * ( i + 3 ) \over 2} - S = {i * ( i + 1 ) \over 2} - S + 2 * i + 3为偶数$$
若 $i$ 为偶数，则为了保证奇偶性，需要 $i + 1$
$$sum(i) - S = {i * ( i + 1 ) \over 2} - S为奇数$$
$${(i + 1） * ( i + 2 ) \over 2} - S = {i * ( i + 1 ) \over 2} - S + i + 1为偶数$$

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