【水模拟】#62 A. Irrational problem_represents the operation of taking the remainder).-优快云博客

本文介绍了一个有趣的数论问题，通过给定的四个不同整数，寻找在特定区间内能使特定函数成立的整数数量。使用模拟暴力的方法进行解答，并提供了示例输入输出。

Little Petya was given this problem for homework:

You are given function $\text{[math]}$ (here $\text{[math]}$ represents the operation of taking the remainder). His task is to count the number of integers x in range [a;b] with property f(x) = x.

It is a pity that Petya forgot the order in which the remainders should be taken and wrote down only 4 numbers. Each of 24 possible orders of taking the remainder has equal probability of being chosen. For example, if Petya has numbers 1, 2, 3, 4 then he can take remainders in that order or first take remainder modulo 4, then modulo 2, 3, 1. There also are 22 other permutations of these numbers that represent orders in which remainder can be taken. In this problem 4 numbers wrote down by Petya will be pairwise distinct.

Now it is impossible for Petya to complete the task given by teacher but just for fun he decided to find the number of integers $\text{[math]}$ with property that probability that f(x) = x is not less than 31.4159265352718281828459045%. In other words, Petya will pick up the number xif there exist at least 7 permutations of numbers p1, p2, p3, p4, for which f(x) = x.

Input

First line of the input will contain 6 integers, separated by spaces: p1, p2, p3, p4, a, b (1 ≤ p1, p2, p3, p4 ≤ 1000, 0 ≤ a ≤ b ≤ 31415).

It is guaranteed that numbers p1, p2, p3, p4 will be pairwise distinct.

Output

Output the number of integers in the given range that have the given property.

Sample test(s)

input
2 7 1 8 2 8

output
0

input
20 30 40 50 0 100

output
20

input
31 41 59 26 17 43

output
9

被这道题笑哭……我还想了一会儿有没有数论上的问题呢，tag还是数论呢……一看数据范围，笑哭……模拟暴力……

#include <iostream>
#include <algorithm>
using namespace std;

int main()
{
	int p1,p2,p3,p4,a,b;
	cin>>p1>>p2>>p3>>p4>>a>>b;
	int cnt;
	for(int i=a;i<=b;i++)
	{
		if(i%p1%p2%p3%p4==i)cnt++;
	}
	cout<<cnt;
    return 0;
}