题目一：输入一个递增排序的数组和一个数字s，在数组中查找两个数，使得它们的和正好是s。如果有多对数字的和等于s，输出任意一对即可。

  思路：定义两个指针，一个放在头部，一个放在尾部，因为是有序的。所以当两个指针所指的数的和大于s时，尾指针向前移，和小于s时，头指针向后移。

#include<iostream>
using namespace std;bool FindNumbersWithSum(int data[], int length, int sum, int* num1, int* num2)
{bool found = false;if (length < 1 || num1 == NULL || num2 == NULL){return found;}int behind = length - 1;int ahead = 0;while (behind > ahead){long long curSum = data[ahead] + data[behind];if (curSum == sum){*num1 = data[ahead];*num2 = data[behind];found = true;// cout << *num1 << " " << *num2 << endl;break;}else if (curSum > sum){behind--;}else{ahead++;}}return found;}void test()
{int data[] = {1, 2, 4, 7, 11};int num1, num2;bool result = FindNumbersWithSum(data, 5, 15, &num1, &num2);if (result){cout << num1 << " " << num2 << endl;}
}int main()
{test();return 0;
}

题目二：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，所以打印出3个连续序列1-5、4-6和7-8。
 
  思路：设两个变量small和big，初始small=1，big=2，有如下几种情况：
  1.small + big < s：big++。会使这个序列变长，比如small=1,big=4，则它打印出来的1-4的值而不是1和4.

2.small + big > s：small++。会缩短序列。

3.small + big = s：打印该序列，并且big++。

#include<iostream>
using namespace std;void PrintContinuousSequence(int small, int big)
{for (int i = small; i <= big; i++){cout << i << " ";}cout << endl;
}void FindContinusSequence(int sum)
{if (sum < 3){return;}int small = 1;int big = 2;int middle = (1 + sum) / 2;int curSum = small + big;while (small < middle)  //  最小值应小于中间值{if (curSum == sum){PrintContinuousSequence(small, big);}while (curSum > sum && small < middle){curSum -= small;small++;if (curSum == sum){PrintContinuousSequence(small, big);}}big++;curSum += big;}}int main()
{FindContinusSequence(1000);return 0;
}