POJ1543 + 1013 解题报告

考察点：枚举法

 

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

 

POJ1543 完美立方

    题目描述：

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)

   思想：只需要把4个参数中的所有值都扫描一遍，看是否符合要求。

Source Code：

//寻找完美立方等式  
#include <iostream>  
using namespace std;  
  
int main()  
{  
    int n;  
    cin >> n;  
    int a, b, c, d;  
    int three[101];  
    for( int i = 0; i < 101; i++ )  
        three[ i ] = i * i *i;  
    int result;  
    for( a = 3; a <= n; a++)  
        for( b = 2; b <=n; b++ )  
            for( c = 2; c <= n; c++ )  
                for( d =2; d <= n; d++ )  
                {  
                    result = ( three[ b ] + three[ c ] + three[ d ] ) ;  
                    if( three[ a ] == result )  
                        cout << "Cube = " << a << ",Triple = (" << b << "," << c << "," << d <<")" << endl;  
                }  
}  

结果：

题目要求输出的时候，b，c，d按照非降序排列输出，则b < c < d,修改代码，提交，AC通过。代码如下：

//寻找完美立方等式  
#include <iostream>  
using namespace std;  
  
int main()  
{  
    int n;  
    cin >> n;  
    int a, b, c, d;  
    int three[101];  
    for( int i = 0; i < 101; i++ )  
        three[ i ] = i * i *i;  
    int result;  
    for( a = 3; a <= n; a++)  
        for( b = 2; b <=n; b++ )  
            for( c = b; c <= n; c++ )  
                for( d =c; d <= n; d++ )  
                {  
                    result = ( three[ b ] + three[ c ] + three[ d ] ) ;  
                    if( three[ a ] == result )  
                        cout << "Cube = " << a << ", Triple = (" << b << "," << c << "," << d <<")" << endl;  
                }  
}  

当然，这样搜索的范围仍然太多，时间复杂度还是太大，我们需要做优化。

 

可以根据已知的东西来推出几个能够缩小范围的条件：

1.a一定大于b，c，d，所以bcd的for循环。只需要小于a就可以了，

2.在4层循环的第二层设立条件，如果 

three[a] < three[b] * 3;  

则没有必要继续向下循环了，因为c，和d向下循环的话只会越来越大。继续修改代码"

//寻找完美立方等式  
#include <iostream>  
using namespace std;  
  
int main()  
{  
    int n;  
    cin >> n;  
    int a, b, c, d;  
    int three[101];  
    for( int i = 0; i < 101; i++ )  
        three[ i ] = i * i *i;  
    int result;  
  
    for( a = 3; a <= n; a++)  
    {  
        for( b = 2; b <a; b++ )  
        {  
            if( three[ a ] < three[ b ] * 3)  
                break;  
            for( c = b; c < a; c++ )  
                for( d =c; d < a; d++ )  
                {  
                    result = ( three[ b ] + three[ c ] + three[ d ] ) ;  
                    if( three[ a ] == result )  
                        cout << "Cube = " << a << ", Triple = (" << b << "," << c << "," << d <<")" << endl;  
                }  
        }  
    }  
}