ZOJ Problem Set - 1331 Perfect Cubes

这个题啊,一看就是暴力，暴力，暴力！！！

但是，暴力也是要做到暴力的恰到好处；看题

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 <= 200.

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.

The first part of the output is shown here:

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)

Note: The programmer will need to be concerned with an efficient implementation. The official time limit for this problem is 2 minutes, and it is indeed possible to write a solution to this problem which executes in under 2 minutes on a 33 MHz 80386 machine. Due to the distributed nature of the contest in this region, judges have been instructed to make the official time limit at their site the greater of 2 minutes or twice the time taken by the judge's solution on the machine being used to judge this problem.

好啦，那就直接暴力吧；直接的代码


#include<iostream>
using namespace std;
int main(){
    int a[3]={0};
    for(int i=6;i<=200;i++){
        for(int j=2;j<=200;j++){
            for(int k=2;k<=200;k++){
                for(int h=2;h<=200;h++){
                    if(i*i*i==j*j*j+k*k*k+h*h*h){
                        cout<<"Cube = "<<i<<","<<" Triple = ("<<j<<","<<k<<","<<h<<")"<<endl;
                    }

                }
            }
        }
    }
    return 0;
}

提交~心情美美哒·······················我擦，Wrong Answer？？？和答案一对比，哦哦哦原来是b,c,d如果符合“完美立方数”然后他们三个的数值的对应关系变了，那么这个答案也不会输出。

哦，那好，这回明白了。明白是明白了，这你妹怎么限制条件啊，不要慌，看AC代码

#include<iostream>
using namespace std;
int main(){
    int a[3]={0};
    for(int i=6;i<=200;i++){
        for(int j=2;j<=200;j++){
            for(int k=j;k<=200;k++){
                for(int h=k;h<=200;h++){
                    if(i*i*i==j*j*j+k*k*k+h*h*h){
                        cout<<"Cube = "<<i<<","<<" Triple = ("<<j<<","<<k<<","<<h<<")"<<endl;
                    }

                }
            }
        }
    }
    return 0;
}

呀，是不明白了！这样就避免了重复值的出现，ok搞定；

就比如6 3 4 5

循环完后 j=4开始循环此时有了k=j;就不可能出现k=3的情况；

ZOJ Problem Set - 1331 Perfect Cubes

猜你喜欢