• OpenJudge 2810(1543) 完美立方 / Poj 1543 Perfect Cubes


    1.链接地址:

    http://bailian.openjudge.cn/practice/2810/

    http://bailian.openjudge.cn/practice/1543/

    http://poj.org/problem?id=1543

    2.题目:

    Perfect Cubes
    Time Limit: 1000MS   Memory Limit: 10000K
    Total Submissions: 13190   Accepted: 6995

    Description

    For hundreds of years Fermat's Last Theorem, which stated simply that for n > 2 there exist no integers a, b, c > 1 such that a^n = b^n + c^n, has remained elusively unproven. (A recent proof is believed to be correct, though it is still undergoing scrutiny.) It is possible, however, to find integers greater than 1 that satisfy the "perfect cube" equation a^3 = b^3 + c^3 + d^3 (e.g. a quick calculation will show that the equation 12^3 = 6^3 + 8^3 + 10^3 is indeed true). This problem requires that you write a program to find all sets of numbers {a,b,c,d} which satisfy this equation for a <= N.

    Input

    One integer N (N <= 100).

    Output

    The output should be listed as shown below, one perfect cube per line, in non-decreasing order of a (i.e. the lines should be sorted by their a values). The values of b, c, and d should also be listed in non-decreasing order on the line itself. There do exist several values of a which can be produced from multiple distinct sets of b, c, and d triples. In these cases, the triples with the smaller b values should be listed first.

    Sample Input

    24

    Sample Output

    Cube = 6, Triple = (3,4,5)
    Cube = 12, Triple = (6,8,10)
    Cube = 18, Triple = (2,12,16)
    Cube = 18, Triple = (9,12,15)
    Cube = 19, Triple = (3,10,18)
    Cube = 20, Triple = (7,14,17)
    Cube = 24, Triple = (12,16,20)

    Source

    3.思路:

    枚举+打表(减少计算次数)

    注意a要升序排列,然后b,c,d再升序排列

    4.代码:

     1 #include <iostream>
     2 #include <cstdio>
     3 
     4 #define START_N 2
     5 
     6 using namespace std;
     7 
     8 int main()
     9 {
    10     int n;
    11     cin>>n;
    12 
    13     int *arr_cube = new int[n];
    14 
    15     int i,j,k,p;
    16     for(i = START_N; i <= n; ++i)
    17     {
    18         arr_cube[i - START_N] = i * i * i;
    19         for(j = START_N; j <= i; ++j)
    20         {
    21             for(k = j; k <= i; ++k)
    22             {
    23                 for(p = k; p <= i; ++p)
    24                 {
    25                     if(arr_cube[i - START_N] == arr_cube[j - START_N]
    26                         + arr_cube[k - START_N] + arr_cube[p - START_N])
    27                     {
    28                         cout<<"Cube = "<<i<<", Triple = ("<<j<<","<<k<<","<<p<<")"<<endl;
    29                     }
    30                 }
    31             }
    32         }
    33 
    34     }
    35 
    36 
    37     delete [] arr_cube;
    38     return 0;
    39 }
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  • 原文地址:https://www.cnblogs.com/mobileliker/p/3546683.html
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