http://noi.openjudge.cn/ch0201/1812/

C++代碼來源于以下鏈接

https://blog.csdn.net/qq_26919935/article/details/77905491

https://blog.csdn.net/xcdq_aaa/article/details/112630852

/* 1812完美立方（2.1基本算法之枚舉） 2.1基本算法之枚舉_1812完美立方 https://blog.csdn.net/qq_26919935/article/details/77905491 http://noi.openjudge.cn/ch0201/1812/ */ #include <iostream> #include<stdio.h> using namespace std; //http://noi.openjudge.cn/ch0201/1812/ //4重循環，關鍵是該break的時候盡早，省了很多無用功 int n,a,b,c,d,a3,b3,c3,d3; int main(int argc, char *argv[]) {cin>>n;for(int a=2;a<=n;a++){a3=a*a*a;for(int b=2;b<=n;b++){b3=b*b*b;if(b3>=a3)break;for(int c=b;c<=n;c++){c3=c*c*c;if(c3+b3>=a3)break;for(int d=c;d<=n;d++){d3=d*d*d;if(d3+c3+b3>a3)break;if(d3+c3+b3==a3){printf("Cube = %d, Triple = (%d,%d,%d)\n",a,b,c,d);}}}}}return 0; }

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/* 2.1基本算法之枚舉_1812完美立方 方法二 http://noi.openjudge.cn/ch0201/1812/https://blog.csdn.net/qq_26919935/article/details/77905491 https://blog.csdn.net/xcdq_aaa/article/details/112630852 */ #include <iostream> using namespace std; int all[20]; int main() {int n;cin >> n;for (int i = 2; i <= n; i++) {for (int j = 2; j <= n; j++) {for (int k = j; k <= n; k++) {for (int l = k; l <= n; l++) {if (i * i * i == j * j * j + k * k * k + l * l * l) {cout << "Cube = " << i << ", Triple = (" << j << ","<< k << "," << l << ")" << endl;}}}}} }

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