群答疑随手记：花椰妹在纸上写下了若干个连续的正整数，她把这些数遮挡住，只告诉蒜头君一共有k个数，它们的和为n,请蒜头君猜猜纸上是哪些数。现在蒜头君向你求助，帮他解决这个问题。（连续整数和）

#include <iostream>
#include <stdio.h>
#include <ctype.h>
#include<algorithm>
#include<cstring> 
#include<stdlib.h>  
using namespace std;
#define MAX 100;
int s2i(string s);
int check(string s);
int check_n(string s);
int check_k(string s);
int check_chinese(string s);
int sum(int i, int j);
int main()
{
    string n0,n1;
    int k, n;
    cout<<"请输入连续正整数的个数k(1<=k<=100):"<<endl;
    cin >> n0;
    while (check(n0) || check_k(n0)) {
        cout << "请输入0-100的正整数，不要输入负数或者小数或字母或汉字或超出范围的数：" << endl;
        cin >> n0;
    }
    cout << "请输入k个连续的正整数和的值n:" << endl;
    cin >> n1;
    while (check(n1)|| check_n(n1)){
        cout << "请输入1<=k<=10^9的正整数，不要输入负数或者小数或字母或汉字或超出范围的数：" << endl;
        cin >> n1;
    }
    k = s2i(n0);
    n = s2i(n1);
    while (true) {
        int i, j;
        char flag = false;
        for (i = 1; i <= n / 2; i++) {
            for (j = i; j < n; j++) {
                if (sum(i, j) == n && (j - i + 1 )==k) {
                    flag = true;
                    for (int p = i; p <= j; p++)
                        cout << p << " ";
                    cout << endl;
                    break;
                }
            }
        }
        if (flag == false)
            cout << "不存在"<<k<<"个连续的正整数和的值为"<<n << endl;
        break;
    }
    return 0;
}
int check_chinese(string s)
{
    string t;
    for (int i = 0; i < s.length(); i++)
    {
        if (s[i] < 255 && s[i]>0) //扩充的ASCII字符范围为0-255,如是,处理一个字节
        {
            t.append(s.substr(i, 1));
            t.append("/");
        }
        else //<0,>255的是汉字,处理两个字节
        {
            t.append(s.substr(i, 2));
            t.append("/");
            ++i;
            return -1;
        }
    }
    return  0;
}
int sum(int i, int j)
{
    return  (i + j) * (j - i + 1) / 2;
}
int s2i(string s)
{
    return atoi(s.c_str());
}
int check(string s) {
    if (check_chinese(s)==-1)
    {
        return -1;
    }
    else
    {
        char* a = new char[s.length()];
        strcpy(a, s.c_str());
        int i;
        for (i = 0; i < s.length(); i++) {
            if (!isdigit(a[i])) {
                return -1;
            }
        }
        return 0;
        delete[]a;
        a = NULL;
    }
}
int check_k(string s)
{
    int k;
    k = s2i(s);//string转为int存储
    if (k < 1 || k>100)
    {
        return -1;
    }
    else
    {
        return 0;
    }
}
int check_n(string s)
{
    int n;
    n = s2i(s);//string转为int存储
    if (n < 1 || n> pow(10, 9))
    {
        return -1;
    }
    else
    {
        return 0;
    }
}