每日编程-20170308-2

题目：给出两个自然数N和n，找到一组连续的自然数，其长度大于等于n，其和等于N。

输入：N（空格）n

输出：自然数列，每个数之间有空格，最后一个数后无空格 | 没有答案则输出：No answer

解答：

#include <iostream>
#include <vector>
using std::cout; using std::endl; using std::cin;
using std::vector;

int main() {

    int N, n, sum = 0;　　//sum为总和
    vector<int> answer;　　//答案以vector形式保存,因为不确定会有多少
    cin >> N >> n;　　//输入参数
    for (auto i = 1; i <= N/2; i++)　　//遍历1到N/2，考虑到(N/2)+(N/2)+1 > N
    {
        for (auto j = i; sum < N ; j++)　　//从i开始遍历，直到数列和不小于N
        {
            sum += j;
            answer.push_back(j);
        }
        if (sum == N && ((answer.end() - answer.begin()) > n))　　//如果sum等于N，并且answer长度大于n
        {
            for (auto beg = answer.begin(); beg != answer.end();beg++)　　//输出answer
            {
                cout << *beg;
                if (beg+1 != answer.end())
                {
                    cout << " ";
                }
            }
            break;
        }
        else　　//如果sum不等于N或者answer长度小于等于n
        {
            answer = {};　　//清空答案，sum清零
            sum = 0;
        }
    }
    if (answer.begin() == answer.end())　　//如果answer为空，没有答案
    {
        cout << "No answer" << endl;
    }
}

 第二版

#include <iostream>
#include <vector>
using std::cout; using std::endl; using std::cin;
using std::vector;

int main() {

    int N, n, sum = 0;
    vector<int> answer;
    cin >> N >> n;
    for (auto i = 1; i <= N/n; i++)　　//数学改进，n(N/n) + n-1 > n
    {
        for (auto j = i; sum < N ; j++)
        {
            sum += j;
            answer.push_back(j);
        }
        if (sum == N && (answer.size() > n)) 　　//刚才脑残了，把size忘了
        {
            for (auto beg = answer.begin(); beg != answer.end();beg++)
            {
                cout << *beg;
                if (beg+1 != answer.end())
                {
                    cout << " ";
                }
            }
            break;
        }
        else
        {
            answer = {};
            sum = 0;
        }
    }
    if (answer.begin() == answer.end())
    {
        cout << "No answer" << endl;
    }
}