Irrational problem-优快云博客

本文链接：https://blog.youkuaiyun.com/clicl/article/details/6269487

作者首次参加Codeforces竞赛并使用STL解决了一个涉及数学和组合问题的任务。任务要求计算特定范围内整数的数量，这些整数在经过一系列取余操作后能保持不变的概率至少达到一定标准。

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第一次做CF，第一次用STL，纪念一下！

Codeforces Beta Round #62

A. Irrational problem

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 x if 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

Input

20 30 40 50 0 100

Output

Input

31 41 59 26 17 43

Output

#include<cstdio> #include<algorithm> #include<cstdlib> #include<iostream> using namespace std; int main(void) { int a,b; int count,p[4]; while(scanf("%d%d%d%d%d%d",&p[0],&p[1],&p[2],&p[3],&a,&b) == 6) { int nums = 0; for(int t = a;t <= b; t++) { count = 0; sort(p,p+4); do { int n = t; for(int i = 0;i < 4;i++) n %= p[i]; if(n == t) count++; }while(next_permutation(p,p+4)); if(count >= 7) nums++; } cout << nums; } return 0; }

比着刘汝佳白皮书上的下一个排列写的，继续努力吧！