// 面试题57（二）：为s的连续正数序列// 题目：输入一个正数s，打印出所有和为s的连续正数序列（至少含有两个数）。// 例如输入15，由于1+2+3+4+5=4+5+6=7+8=15，所以结果打印出3个连续序列1～5、// 4～6和7～8。#include 
     
      void PrintContinuousSequence(int small, int big);void FindContinuousSequence(int sum){    if (sum < 3)        return;    int small = 1;    int big = 2;    int middle = (1 + sum) / 2;//因为序列要求至少包含两个数字，所以最大不会超过(1 + sum) / 2    int curSum = small + big;    while (small < middle)    {        if (curSum == sum)            PrintContinuousSequence(small, big);        while (curSum > sum && small < middle)//如果大于要求值，就减去最小值，并把最小值加一        {            curSum -= small;            small++;            if (curSum == sum)                PrintContinuousSequence(small, big);        }        big++;//然后就会出现小于要求值的情况，就加上++big        curSum += big;    }}void PrintContinuousSequence(int small, int big)//打印序列{    for (int i = small; i <= big; ++i)        printf("%d ", i);    printf("\n");}// ====================测试代码====================void Test(const char* testName, int sum){    if (testName != nullptr)        printf("%s for %d begins: \n", testName, sum);    FindContinuousSequence(sum);}int main(int argc, char* argv[]){    Test("test1", 1);    Test("test2", 3);    Test("test3", 4);    Test("test4", 9);    Test("test5", 15);    Test("test6", 100);    system("pause");    return 0;}