zoukankan      html  css  js  c++  java
  • HDUSTOJ-1559 Vive la Difference!(简单题)

    1559: Vive la Difference!

    时间限制: 3 Sec  内存限制: 128 MB
    提交: 18  解决: 14
    [提交][状态][讨论版]

    题目描述

    Take any four positive integers: a, b, c, d. Form four more, like this:

    | a - b|,| b - c|,| c - d|,| d - a|

    That is, take the absolute value of the differences of a with b, b with c, c with d, and d with a. (Note that a zero could crop up, but they'll all still be non-negative.) Then, do it again with these four new numbers. And then again. And again. Eventually, all four integers will be the same. For example, start with 1,3,5,9:


    1 3 5 9 
    2 2 4 8 (1) 
    0 2 4 6 (2) 
    2 2 2 6 (3) 
    0 0 4 4 (4) 
    0 4 0 4 (5) 
    4 4 4 4 (6)

    In this case, the sequence converged in 6 steps. It turns out that in all cases, the sequence converges very quickly. In fact, it can be shown that if all four integers are less than 2n, then it will take no more than 3*n steps to converge!

    Given a, b, c and d, figure out just how quickly the sequence converges.

    输入

    There will be several test cases in the input. Each test case consists of four positive integers on a single line ( 1$ le$a, b, c, d$ le$2, 000, 000, 000), with single spaces for separation. The input will end with a line with four 0's.

    输出

    For each test case, output a single integer on its own line, indicating the number of steps until convergence. Output no extra spaces, and do not separate answers with blank lines.

    样例输入

    1 3 5 9
    4 3 2 1
    1 1 1 1
    0 0 0 0
    

    样例输出

    6
    4
    0
    这道题原来以为直接搞会超时的,结果没超时,而且还很充裕
    #include<algorithm>
    #include<iostream>
    #include<cstring>
    #include<cstdio>
     
    int main(){
        int a, b, c, d;
        while(scanf("%d %d %d %d", &a, &b, &c, &d) == 4 && (a + b + c + d)){
            int cnt = 0;
            while(true){
                if(a == b && b == c && c ==d) break;
                int tmp = a;
                a = abs(a - b);
                b = abs(b - c);
                c = abs(c - d);
                d = abs(d - tmp);
                cnt++;
            }
            printf("%d
    ", cnt);
        }
    }
  • 相关阅读:
    STL容器内数据删除
    grep 同时满足多个关键字和满足任意关键字
    程序运行栈空间不足程序崩溃问题
    VS2010中设置程序以管理员身份运行
    python 包详解
    select 详解
    Hdu 1166
    CF1204C
    CF1204B
    CF1204A
  • 原文地址:https://www.cnblogs.com/Pretty9/p/7406638.html
Copyright © 2011-2022 走看看