[CSP-J 2022] 解密

思路：

 

AC code:

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#include <iostream> using namespace std; #define int long long   signed main() {     int t;     cin >> t;     while(t--)     {         int n,d,e;         cin >> n >> d >>e ;         int t = n - e * d + 2;         int l = 1,r = t / 2 + 1;//只用到它的一半即可         bool flag = false;         while(l <= r)         {             int mid =(l+ r) >> 1;             int tt = t - mid;             if(mid *tt == n)             {                 cout << mid << " " << t - mid << endl;//如果可以的话能够中途输出的                 flag = true;                 break;             }             if((mid) * (tt) < n)             {                 l = mid + 1;             }             else             {                 r = mid - 1;             }         }         if(flag)         {             continue;         }         if((l) * (t - l) == n)cout << l << " " << t - l << endl;         else cout << "NO" << endl;//判读无解     }     return 0;  }

　　

另外一种解：

 

AC code:

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#include <iostream> #include <cmath> using namespace std; #define int long long   signed main() {     int t;     cin >> t;     while(t--)     {         int n,d,e;         cin >> n >> d >> e;         int p = n - e * d + 2 + sqrt((n - e * d + 2) * (n - e * d + 2) - 4 * n);         int q = n - e * d + 2 - sqrt((n - e * d + 2) * (n - e * d + 2) - 4 * n);         if(p & 1 ||q & 1 || p < 0 || q < 0)//判断是不是正整数         {             cout << "NO" << endl;         }         else         {             cout << min(p / 2,q / 2)  << " " << max(p / 2,q / 2) << endl;         }     }     return 0; }