UVa OJ - 10193 All You Need Is Love

The following program is my ACcepted code for UVA-10193 .
It's a for everybody to learn and discuss.
If there is any mistake or comment, please let me know.  :D

此乃UAV 10193 的AC code!

歡迎一同討論學習，如有錯誤與任何建議請留言 : )

點這裡看題目 Click here to see this Problem!

	//This program is for UVA 10193 All You Need Is Love
	//http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=13&page=show_problem&problem=1134
	/*
	S is made of Love when "S can be divisible by L." (with no remainder)
	When (S1%L == 0 && S2%L == 0), we have "All you need is love!"
	otherwise, we get "Love is not all you need!"
	we want "(S1%L == 0 && S2%L == 0)" is true.
	So we have to find a L, which is Common Divisor of S1 and S2.
	And this L is not 1 !
	*/
	#include<stdio.h>
	#include<string.h>
	// convert a binary number to decimal
	int binaryToDecimal(char a[])
	{
	int len = strlen(a);
	int m =0;
	m = a[0] - '0';// change the character to integer
	for(int i = 1; i < len; i++)
	{
	m *= 2;
	m += a[i] - '0';// change the character to integer
	}
	//printf("str: %s m: %d\n", a, m);// for check
	return m;
	}
	/* find the Greatest Common Divisor
	The parameter a must be bigger than b!
	use the recursive to find GCD.
	*/
	int GCD(int a, int b)
	{
	if(b > a)
	return GCD(b, a);
	if(b == 0)
	return a;
	else
	return GCD(b, a%b);
	}
	int main()
	{
	int n, x, y, gcd;
	char s1[32], s2[32];
	scanf("%d\n", &n);
	for(int t = 1; t <= n; t++)
	{
	gets(s1);
	gets(s2);
	x = binaryToDecimal(s1);
	y = binaryToDecimal(s2);
	gcd = GCD(x, y);
	if(gcd != 1)
	printf("Pair #%d: All you need is love!\n", t);
	else
	printf("Pair #%d: Love is not all you need!\n", t);
	}
	return 0;
	}

view raw UVA 10193 All You Need Is Love.cpp hosted with ❤ by GitHub

Please feel free to use it after adding this blog as an reference. (http://autekroy.blogspot.tw) If there is any mistake or comment, please let me know. :D 

歡迎使用與分享任何內容，但請記得標示此部落格為出處。(http://autekroy.blogspot.tw/) 如果有發現任何的錯誤與建議請留言或跟我連絡。 : )