X = a0 + a1B + a2B2 + ...+ anBn. (其实是B^2。。。。这个坑人)满足复数除法的剪枝真的是强大啊枚举每一个ai, 然后 除复复杂的黑猫, 这里剪枝，判断是否可以整除，复数的除法公式自己推/*ID: meixiny1PROG: testLANG: C++11*/#include <cstdio>#include <cstdlib>#include <iostream>#include <algorithm>#include <cstring>#include <climits>#include <string>#include <vector>#include <cmath>#include <stack>#include <queue>#include <set>#include <map>#include <sstream>#include <cctype>using namespace std;typedef long long ll;typedef pair<int ,int> pii;#define MEM(a,b) memset(a,b,sizeof a)#define CLR(a) memset(a,0,sizeof a);const int inf = 0x3f3f3f3f;const int MOD = 1e9 + 7;#define PI 3.1415926535898//#define LOCALll limit,ok;ll xr,xi,c,d;ll ans;int dfs(ll xr,ll xi,int deep){ ll a ,b = xi,x,y; int check = 0; if(deep>100)return 0; if(xi==0 && xr<=limit && ((xr>=0 && deep == 0 ) zzdcdqzzdcdq (xr>0))){ ok = 1; printf("%lld",xr); return 1; } for(int i=0;i<=limit;i++){ a =xr - i; x = (a*c+d*b)%ans; y = (c*b-a*d)%ans; if(x zzdcdqzzdcdq y)continue; x=(a*c+d*b)/ans; y=(c*b-a*d)/ans; check zzdcdq= dfs(x,y,deep+1); if(check){ printf(",%d",i); return 1; } } return 0;}int main(){#ifdef LOCAL freopen("in.txt", "r", stdin);// freopen("out.txt","w",stdout);#endif int t;scanf("%d",&t); while(t--){ ok = 0; scanf("%lld%lld%lld%lld",&xr,&xi,&c,&d); ans = c*c+d*d; if(sqrt(ans)==(ll)sqrt(ans))limit = (ll)sqrt(ans)-1; else limit = (ll)sqrt(ans); dfs(xr,xi,0); if(ok)printf("n"); else printf("The code cannot be decrypted.n"); } return 0;}