JZ-C-41-Plus

剑指offer第四十一题-拓展：和为s的连续正数序列：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）

//============================================================================
// Name        : JZ-C-41-Plus.cpp
// Author      : Laughing_Lz
// Version     :
// Copyright   : All Right Reserved
// Description : 和为s的连续正数序列：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）
//============================================================================

#include <iostream>
#include <stdio.h>
using namespace std;

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;
    int curSum = small + big; //这里curSum其实是从small到big之间所有数字的累加和

    while (small < middle) {
        if (curSum == sum)
            PrintContinuousSequence(small, big);

        while (curSum > sum && small < middle) {
            curSum -= small; //每次更新curSum值
            small++;

            if (curSum == sum)
                PrintContinuousSequence(small, big);
        }
//        while (curSum <= sum && small < middle) { //自己写的还是不如源码啊。。清晰却不简洁
//            big++;
//            curSum += big;
//            if (curSum >= sum)
//                break;
//        }
        big++;
        curSum += big; //每次更新curSum值
    }
}

void PrintContinuousSequence(int small, int big) {
    for (int i = small; i <= big; ++i)
        printf("%d ", i);

    printf("\n");
}

// ====================测试代码====================
void Test(char* testName, int sum) {
    if (testName != NULL)
        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);

    return 0;
}