给定总和S和整数N，任务是从1到N除去两个连续的整数，以使总和等于S。
例子：
Input: N = 4, S = 3
Output: Yes
sum(1, 2, 3, 4) = 10,
remove 3 and 4 from 1 to N
now sum(1, 2) = 3 = S
Hence by removing 3 and 4 we got sum = S
Input: N = 5, S =3
Output: No
Its not possible to remove two elements
from 1 to N such that new sum is 3
- 方法1：
- 创建一个数组，其元素的范围是1到N。
- 每次删除两个连续的元素，并检查N个自然数之和与两个删除的元素之和之间的差为S。
- 可以使用公式计算N个自然数的总和
sum = (n)(n + 1) / 2
Time complexity: O(n) Space complexity: O(n) - 方法2：
- 我们知道，可以使用公式计算N个自然数的总和
sum = (n)(n + 1) / 2 - 根据问题陈述，我们需要从1到N中删除两个整数，以使剩余整数之和为S。
- 令要删除的两个连续整数为i和i + 1 。
- 所以，
required sum = S = (n)(n + 1) / 2 - (i) - (i + 1) S = (n)(n + 1) / 2 - (2i - 1) Therefore, i = ((n)(n + 1) / 4) - ((S + 1) / 2) - 因此找到我使用上面的公式
- 如果我是整数，则答案为是，否则为否
下面是上述方法的实现：
C++
// C++ program remove two consecutive integers // from 1 to N to make sum equal to S #includeusing namespace std; // Function to find the numbers // to be removed float findNumber(int N, int S) { // typecast appropriately // so that answer is float float i = (((float)(N) * (float)(N + 1)) / 4) - ((float)(S + 1) / 2); // return the obtained result return i; } void check(int N, int S) { float i = findNumber(N, S); // Convert i to integer int integerI = (int)i; // If i is an integer is 0 // then answer is Yes if (i - integerI == 0) cout << "Yes: " << integerI << ", " << integerI + 1 << endl; else cout << "No" << endl; } // Driver code int main() { int N = 4, S = 3; check(N, S); N = 5, S = 3; check(N, S); return 0; }
Java
// Java program remove two consecutive integers // from 1 to N to make sum equal to S class GFG { // Function to find the numbers // to be removed static float findNumber(int N, int S) { // typecast appropriately // so that answer is float float i = (((float)(N) * (float)(N + 1)) / 4) - ((float)(S + 1) / 2); // return the obtained result return i; } static void check(int N, int S) { float i = findNumber(N, S); // Convert i to integer int integerI = (int)i; // If i is an integer is 0 // then answer is Yes if (i - integerI == 0) System.out.println("Yes: " + integerI + ", " + (integerI + 1)) ; else System.out.println("No"); } // Driver code public static void main (String[] args) { int N = 4, S = 3; check(N, S); N = 5; S = 3; check(N, S); } } // This code is contributed by AnkitRai01
Python3
# Python3 program remove two consecutive integers # from 1 to N to make sum equal to S # Function to find the numbers # to be removed def findNumber(N, S) : # typecast appropriately # so that answer is float i = (((N) * (N + 1)) / 4) - ((S + 1) / 2); # return the obtained result return i; def check(N, S) : i = findNumber(N, S); # Convert i to integer integerI = int(i); # If i is an integer is 0 # then answer is Yes if (i - integerI == 0) : print("Yes:", integerI, ",", integerI + 1); else : print("No"); # Driver code if __name__ == "__main__" : N = 4; S = 3; check(N, S); N = 5; S = 3; check(N, S); # This code is contributed by AnkitRai01
C#
// C# program remove two consecutive integers // from 1 to N to make sum equal to S using System; class GFG { // Function to find the numbers // to be removed static float findNumber(int N, int S) { // typecast appropriately // so that answer is float float i = (((float)(N) * (float)(N + 1)) / 4) - ((float)(S + 1) / 2); // return the obtained result return i; } static void check(int N, int S) { float i = findNumber(N, S); // Convert i to integer int integerI = (int)i; // If i is an integer is 0 // then answer is Yes if (i - integerI == 0) Console.WriteLine("Yes: " + integerI + ", " + (integerI + 1)) ; else Console.WriteLine("No"); } // Driver code public static void Main() { int N = 4, S = 3; check(N, S); N = 5; S = 3; check(N, S); } } // This code is contributed by AnkitRai01
输出：Yes: 3, 4 No时间复杂度： O(1)
空间复杂度： O(1) - 我们知道，可以使用公式计算N个自然数的总和