Iahub and his friend Floyd have started painting a wall. Iahub is painting the wall red and Floyd is painting it pink. You can consider the wall being made of a very large number of bricks, numbered 1, 2, 3 and so on.
Iahub has the following scheme of painting: he skips x?-?1 consecutive bricks, then he paints the x-th one. That is, he‘ll paint bricks x, 2·x, 3·x and so on red. Similarly, Floyd skips y?-?1 consecutive bricks, then he paints the y-th one. Hence he‘ll paint bricks y, 2·y, 3·y and so on pink.
After painting the wall all day, the boys observed that some bricks are painted both red and pink. Iahub has a lucky number a and Floyd has a lucky number b. Boys wonder how many bricks numbered no less than a and no greater than b are painted both red and pink. This is exactly your task: compute and print the answer to the question.
The input will have a single line containing four integers in this order: x, y, a, b. (1?≤?x,?y?≤?1000, 1?≤?a,?b?≤?2·109,a?≤?b).
Output a single integer — the number of bricks numbered no less than a and no greater than b that are painted both red and pink.
2 3 6 18
3
这是一道简单题，也隔了一段时间没做简单题目了。
这次感觉又不一样了，可以很快就能写出很优雅的代码了，故此很想贴贴自己的代码。
优雅代码的关键就是要利用数学的思想去解：
本题的实质是可以转化为求最大公倍数的的问题，然后利用Inclusion-exclusion（包含和不包含）的原则，计算有多少个数能被a除尽这个公倍数，有多少个数能被b除尽这个公倍数，然后相减就得到最终答案了。
#include <stdio.h>
using namespace std;
int TheWallGCD(int a, int b)
{
while (b)
{
int t = b;
b = a % b;
a = t;
}
return a;
}
void TheWall340A()
{
int x, y, a, b;
scanf("%d %d %d %d", &x, &y, &a, &b);
int r = TheWallGCD(x, y);
r = x / r * y;
int m = a % r? 0 : 1;
printf("%d", b / r - a / r + m);
}
codeforces The Wall - 题解,布布扣,bubuko.com
原文地址：http://blog.csdn.net/kenden23/article/details/27544033
