这道题明显是一道暴力。
暴力枚举每一个 (a, b, c, d)
所以我就写了一个暴力。每个 (a, b, c, d) 都从 (1) 枚举到 (100)
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#define line cout << endl
using namespace std;
int lf(int x) {
return x * x * x;
}
int main() {
for (int a = 1; a <= 100; a++) {
for (int b = 1; b <= 100; b++) {
for (int c = 1; c <= 100; c++) {
for (int d = 1; d <= 100; d++) {
if (lf(a) == lf(b) + lf(c) + lf(d)) {
cout << "Cube = " << a << ", Triple = (" << b << "," << c << "," << d << ")";
line;
}
}
}
}
}
return 0;
}
结果,成功的 WA 掉了(((
为什么?
我们仔细观察题目给出的前几个样例:
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)
而再看上面的程序输出的前几个:
Cube = 6, Triple = (3,4,5)
Cube = 6, Triple = (3,5,4)
Cube = 6, Triple = (4,3,5)
Cube = 6, Triple = (4,5,3)
Cube = 6, Triple = (5,3,4)
Cube = 6, Triple = (5,4,3)
Cube = 9, Triple = (1,6,8)
Cube = 9, Triple = (1,8,6)
Cube = 9, Triple = (6,1,8)
Cube = 9, Triple = (6,8,1)
Cube = 9, Triple = (8,1,6)
Cube = 9, Triple = (8,6,1)
发现什么问题了吗? 相信 全谷最聪明的你 已经发现了问题。
仔细观察 Cube = 6
的情况:
我的程序:
Cube = 6, Triple = (3,4,5)
Cube = 6, Triple = (3,5,4)
Cube = 6, Triple = (4,3,5)
Cube = 6, Triple = (4,5,3)
Cube = 6, Triple = (5,3,4)
Cube = 6, Triple = (5,4,3)
正确答案:
Cube = 6, Triple = (3,4,5)
有可能题目没有翻译清晰
发现每个 Cube
只需要输出一组,那么我们就可以改变一下循环的初始状态。
将 (a,b) 从 (2) 到 (100) 枚举,将 (c) 从 (b) 到 (100) 枚举,将 (d) 从 (c) 到 (100) 枚举。
最后放上代码:
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#define line cout << endl
using namespace std;
int lf(int x) {//立方的函数
return x * x * x;
}
int main() {
// freopen("1.out", "w", stdout);
for (int a = 2; a <= 100; a++) {
for (int b = 2; b <= 100; b++) {
for (int c = b; c <= 100; c++) {
for (int d = c; d <= 100; d++) {
if (lf(a) == lf(b) + lf(c) + lf(d)) {
cout << "Cube = " << a << ", Triple = (" << b << "," << c << "," << d << ")";
line;
}
}
}
}
}
return 0;
}