/*题目1：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。题目2：输入一个正数，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 <iostream>using namespace std;//和为Sum的两个数字bool FindTwoNumbersWithSum(int data[], int nlength, int &num1, int &num2, int Sum){ bool found = false; if(data == NULL || nlength < 1) return found; int aHead = 0; int Behind = nlength - 1; while(aHead < Behind) { int curSum = data[aHead] + data[Behind]; if(curSum == Sum) { num1 = data[aHead]; num2 = data[Behind]; found = true; break; }else if(curSum < Sum) aHead++; else --Behind; } return found;}//输出连续数字void PrintContinousSequence(int small, int big){ while(small <= big) { cout << small << " "; ++small; } cout << endl;}//和为Sum的连续整数序列void FindNumbersWithContinuousSequence(int Sum){ if(Sum < 3) return; //初始化 int small = 1; int big = 2; int curSum = small + big; int middle = (Sum + 1) / 2; while(small < middle) { if(curSum == Sum) PrintContinousSequence(small, big); while(curSum < Sum && small < middle) { big++; curSum += big; if(curSum == Sum) PrintContinousSequence(small, big); } curSum -= small; small++; }}//=======================测试代码====================void Test(char *TestName, int data[], int nlength, int Sum, bool Expected){ if(TestName != NULL) cout << TestName << " Begins:" << endl; int num1 = 0; int num2 = 0; bool Result = FindTwoNumbersWithSum(data, nlength, num1, num2, Sum); if(Result == Expected) { if(Result) { if(num1 + num2 == Sum) cout << "Passed!" << endl; else cout << "Failed!" << endl; }else cout << "Passed!" << endl; } else cout << "Failed!" << endl;}//=======================测试代码B===========================void TestB(char *TestName, int Sum){ if(TestName != NULL) cout << TestName << " for " << Sum << " Begins:" << endl; FindNumbersWithContinuousSequence(Sum);}//=====================测试用例========================//存在和为Sum的数对，且在数组中间void Test1(){ int data[] = {1, 2, 4, 7, 11, 15}; Test("Test1", data, sizeof(data) / sizeof(int), 15, true);}//存在和为Sum的数对且在数组两端void Test2(){ int data[] = {1, 2, 4, 7, 11, 16}; Test("Test2", data, sizeof(data) / sizeof(int), 17, true);}//不存在和为Sum的数对void Test3(){ int data[] = {1, 2, 4, 7, 11, 16}; Test("Test3", data, sizeof(data) / sizeof(int), 10, false);}//鲁棒性测试void Test4(){ Test("Test4", NULL, 0, 10, false);}int main(){ Test1(); Test2(); Test3(); Test4(); TestB("Test1", 1); TestB("Test2", 3); TestB("Test3", 4); TestB("Test4", 9); TestB("Test5", 15); TestB("Test6", 100); return 0;}
