How to use c + + to determine leap year, it is important that you enter illegal is output error. Behind this condition, how to write for a great god

CodePudding user response:

If (input year for non integer)
cout<" The Error "& lt; CodePudding user response:

int year;
While (1) {
Int ok=the scanf (" % d ", & amp; Year);
If (1!=ok | | year & lt; 1000 | | year & gt; 3000) {
Printf (" Error: must be greater than 1000 year is less than 3000, please input again \ n ");
The fflush (stdin);
}
}

CodePudding user response:

What input illegal means, the Numbers?
Can use a regular, for example
STD: : regex e (" \ \ d {4} ");
STD: : string STR.
While (1) {
Printf (" do input year: ");
STD: : cin> STR;
Int ok=STD: : regex_match (STR, e);
If (ok) break;
Printf (" error, both please input again. ");
}

CodePudding user response:

If (y>=0 & amp; & (y %==0 400 | | (y % 4==0 & amp; & Y % 100!
=0)))
This is a leap year judgment condition

The else

cout<" The error "

This is the meaning of

CodePudding user response:

1!=the scanf is determined, and required to enter an integer but enter
the condition of the floating point Numbers or lettersThe year behind the range is judged by effective range, for 3 years BC (3) is a leap year, when asked in 10000 is a leap year please

Common is 4, one hundred, 400 three judgments, and indeed in 3200, at the end of 172800 judgment?
The expression of the uncertainty, which is a leap year younger sister method you write