#include<bits/stdc++.h> using namespace std; int main(){ int a,b,c,d; }

不知道做过多少遍 上代码 一年之后就会变成远古题解了罢

跟传统的水仙花数不太一样，多了YES和NO

老样子，附注释： qwq写注释真的很累，希望大家点上一个大大的赞，这是我更新的动力~

#include <iostream> #include <cmath> using namespace std; int main(){ int n,a,b,c; cin>>n; a = n/100; b= n%100/10; c = n%100%10; a = pow(a,3); b= pow(b,3); c = pow(c,3); if(a+b+c == n) cout<<"YES"; else cout<<"NO"; return 0; }

————###禁止白嫖###———— ————L(' v' )/ +U!—————— ———————————————— ———————————————— ————###题解如下###————— #include<bits/stdc++.h> using namespace std; int main(){ int a,b,c,d; cin>>a; b=a/100; c=a/10%10; d=a%10; if(pow(b,3)+pow(c,3)+pow(d,3)==a){ cout<<"YES"; } else{ cout<<"NO";

#include<bits/stdc++.h> using namespace std; int main(){ int a,b,c,d; cin>>a; b=a/100; c=a/10%10; d=a%10; if(pow(b,3)+pow(c,3)+pow(d,3)==a){ cout<<"YES"; } else{ cout<<"NO"; } return 0; }

#include<iostream> #include<cmath> using namespace std; int main(){ int n; cin>>n; if(pow(n/100,3)+pow(n/10%10,3)+pow(n%10,3)==n){ cout<<"YES"; } else{ cout<<"NO"; } return 0; }

这里有pow会比较简单

#include <bits/stdc++.h> using namespace std; int main(){ int a; cin>>a; int g,s,b; g=a%10; s=a/10%10; b=a/100; if(pow(g,3)+pow(s,3)+pow(b,3)==a){ cout<<"YES"; }else{ cout<<"NO"; } return 0; }

#include <bits/stdc++.h> using namespace std; int main(){ int a,a1,b1,c1; cin >> a; a1 = a%10; b1 = a/10%10; c1 = a/100; if(a1a1a1 + b1b1b1 + c1c1c1 == a){ cout << "YES"; } else { cout << "NO"; } return 0; }

#include<bits/stdc++.h> using namespace std; int main(){ int n; cin>>n; int a,b,c; a=n/100; b=n%100/10; c=n%10; if(aaa+bbb+ccc==n){ cout<<"YES"; } else cout<<"NO"; return 0; }

#include <iostream> using namespace std; int main(){ int a,x,y,z; cin>>a; x=a%10; y=a/10%10; z=a/100; if (xxx+yyy+zzz==a){ cout<<"YES"; } else{ cout<<"NO"; } return 0; }

#include <bits/stdc++.h> using namespace std; int main(){ int a; cin>>a; int g,s,b; g=a%10; s=a/10%10; b=a/100; if(pow(g,3)+pow(s,3)+pow(b,3)==a){ cout<<"YES"; }else{ cout<<"NO"; } return 0; }

#include<bits/stdc++.h> using namespace std; int main(){ int a,b,c,d; cin>>a; b=a/100; c=a/10%10; d=a%10; if(pow(b,3)+pow(c,3)+pow(d,3)==a){ cout<<"YES"; } else{ cout<<"NO"; } return 0; } 记得点赞