You are given an year Y, find the next identical calendar year to Y.

Examples:

Input : 2017 Output : 2023 Input : 2018 Output : 2029

An year** x** is identical to a given previous year **y** if following two conditions are satisfied.

- x starts with same day as y.
- If y is leap year, then x is also. If y is not leap year, then x is also not.

The idea is to check all years one by one (starting from next year). We keep track of number of days moved ahead. If total moved days is 7, then current year begins with same day. We also check if leap-ness of current year is same as y. If both conditions are satisfied, we return current year.

// C++ program to find next identical year #include<iostream> using namespace std; // Function for finding extra days of year // more than complete weeks int extraDays(int y) { // If current year is a leap year, then // it number of weekdays move ahead by // 2 in terms of weekdays. if (y%400==0 || y%100!=0 && y%4==0) return 2; // Else number of weekdays move ahead // by 1. return 1; } // Returns next identical year. int nextYear(int y) { // Find number of days moved ahead by y int days = extraDays(y); // Start from next year int x = y + 1; // Count total number of weekdays // moved ahead so far. for (int sum=0; ; x++) { sum = (sum + extraDays(x)) % 7; // If sum is divisible by 7 and leap-ness // of x is same as y, return x. if ( sum==0 && (extraDays(x) == days)) return x; } return x; } // driver program int main() { int y = 2018; cout << nextYear(y); return 0; }

Output:

2029

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