Codeforces Round 1046 (Div. 2)——D. For the Champion · znzryb
## 思路讲解
这种题目就是要多想，多写，想办法去除绝对值。
把式子写下来。
```cpp
ll dis=ask('U',mod);
sort(all(A),[&](Point a,Point b){
return 2*2*mod-a.x()-a.y() < 2*2*mod-b.x()-b.y();
});
Point a=A[0];
// 知道了x+y是多少
// mod*2+X+mod*2+Y-a.x()-a.()=dis
ll XY=dis-2*2*mod+a.x()+a.y();
// #ifdef LOCAL
// debug(XY);
// #endif
ask('D',mod);
ask('D', mod);
ask('D',mod);
ll dis2=ask('D', mod);
sort(all(A),[&](Point a,Point b){
return 2*2*mod-a.x()+a.y() < 2*2*mod-b.x()+b.y();
});
a=A[0];
// X+2*mod-x+y-(Y-2*mod)=dis2
// dis2+x-y-2*2*mod
ll X_Y=dis2+a.x()-a.y()-2*2*mod;
```
## AC代码
[https://codeforces.com/contest/2136/submission/336079080](https://codeforces.com/contest/2136/submission/336079080)
::: 源代码
```cpp
// by Gospel_rock
#include <algorithm>
#include <bits/stdc++.h>
#define all(vec) vec.begin(),vec.end()
#define PB push_back
#define EB emplace_back
#define EM emplace
#define FI first
#define SE second
#define pct __builtin_popcountll
#define ctz __builtin_ctzll
#define SZ(a) ((long long) a.size())
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (int i = (a); i >= (b); --i)
#define debug(var) cerr << #var <<":"<<var<<"\n";
#define cend cerr<<"\n-----------\n"
#define fsp(x) fixed<<setprecision(x)
#define minn(vec) *min_element(vec.begin(),vec.end())
#define maxx(vec) *max_element(vec.begin(),vec.end())
using namespace std;
#ifdef LOCAL
template<typename T> void _print(T x) { cerr << x; }
// 可以打印多层数组，但是格式不是很好
template<typename T> void _print(vector<T> v) { cerr << "[ "; for (T i : v) _print(i), cerr << " "; cerr << "]"; }
template<typename T> void _print(set<T> s) { cerr << "{ "; for (T i : s) _print(i), cerr << " "; cerr << "}"; }
template<typename T, typename U> void _print(pair<T, U> p) { cerr << "{"; _print(p.first); cerr << ", "; _print(p.second); cerr << "}"; }
template<typename T, typename U> void _print(map<T, U> m) { cerr << "{ "; for (auto &p : m) _print(p), cerr << " "; cerr << "}"; }
// 打印两层数组，格式好一些
template<typename T> void _print(vector<vector<T>> v){cerr<<"{\n";for (auto i : v) _print(i), cerr << "\n";cerr<<" }";}
#endif
using ll = long long;using ull = unsigned long long;
using ld=double;using LD=long double;
using i128 = __int128;
using pdd = pair<ld,ld>;using plb = pair<ll,bool>;
using pll = pair<ll,ll>;using vll=vector<ll>;using vb=vector<bool>;
using tll = tuple<ll,ll,ll>;using vii=vector<int>;
using vpll = vector<pll>;using vtll = vector<tll>;
using vdb = vector<ld>;using vc=vector<char>;
using arr3 = array<ll,3> ;using arr2 = array<ll,2>;
using CD=complex<double>;
static constexpr ll MAXN=(ll)1e6+10,INF=(ll)1e17+3; // 1e17+3是素数 1e18+9是素数
static constexpr ll mod=(ll)1e9; // 998244353;
static constexpr double eps=1e-8;const double pi=acos(-1.0);
ll N,M,Q,X,K,T,lT;
inline void __();
signed main()
{
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
cin>>lT;
while(lT--)
__();
return 0;
}
/*
*/
using Real = ll;
using Point = complex<Real>;
#define x() real()
#define y() imag()
Real cross(const Point &a, const Point &b) {
return (conj(a) * b).imag();
}
Real dot(const Point &a, const Point &b) {
return (conj(a) * b).real();
}
Real mandis(const Point &a,const Point &b){
return abs(a.x()-b.x())+abs(a.y()-b.y());
}
int sgn(Real x) {
if (-eps < x && x < eps) return 0;
return x < 0 ? -1 : 1;
}
namespace std
{
bool operator <(const Point& a, const Point& b){
return sgn(real(a) - real(b)) ? real(a) < real(b) : imag(a) <
imag(b);
}
bool operator ==(const Point& a, const Point& b){
if(!sgn(a.imag()-b.imag()) && !sgn(a.real()-b.real())){
return true;
}
return false;
}
bool operator !=(const Point& a, const Point& b){
if(!sgn(a.imag()-b.imag()) && !sgn(a.real()-b.real())){
return false;
}
return true;
}
istream& operator>>(istream& in, Point& c) {
Real r, i;
in >> r >> i;
c = Point(r, i);
return in;
}
}
// 采用两点式
struct Line{
Point a,b;
Line(){}
Line(Point a,Point b):a(a),b(b) {}
bool operator<(const Line&other) const{
return a!=other.a?a<other.a:b<other.b;
}
bool operator==(const Line&other) const{
return a==other.a && b==other.b;
}
};
istream& operator>>(istream& in, Line& e) {
Real a,b,c,d;
in >> a >> b >> c >> d;
e = Line(Point(a,b),Point(c,d));
return in;
}
// p+tv=q+sw
// (p-q) x w+tv x w = 0
// t=-(p-q)x w/(vxw)
// 传入的是方向式 拿到的是两直线交点
Point getintersect(const Line &a,const Line &b){
Point p=a.a, v=a.b-a.a,q=b.a,w=b.b-b.a;
Point u=p-q;
Real t=cross(w,u)/cross(v,w);
return p+v*t;
}
bool isseginterline(Line seg,Line e){
if(sgn(cross(e.a-seg.a, e.b-seg.a))*sgn(cross(e.a-seg.b, e.b-seg.b))>0){
return false;
}
return true;
}
ll ask(char a,ll b){
cout<<"? "<<a<<" "<<b<<endl;
ll res;cin>>res;
return res;
}
inline void __(){
cin>>N;
if(N==-1) exit(0);
vector<Point> A;
FOR(i,1,N){
Point a;cin>>a;
A.EB(a);
}
ask('R',mod);
ask('R',mod);
ask('U',mod);
ll dis=ask('U',mod);
sort(all(A),[&](Point a,Point b){
return 2*2*mod-a.x()-a.y() < 2*2*mod-b.x()-b.y();
});
Point a=A[0];
// 知道了x+y是多少
// mod*2+X+mod*2+Y-a.x()-a.()=dis
ll XY=dis-2*2*mod+a.x()+a.y();
// #ifdef LOCAL
// debug(XY);
// #endif
ask('D',mod);
ask('D', mod);
ask('D',mod);
ll dis2=ask('D', mod);
sort(all(A),[&](Point a,Point b){
return 2*2*mod-a.x()+a.y() < 2*2*mod-b.x()+b.y();
});
a=A[0];
// X+2*mod-x+y-(Y-2*mod)=dis2
// dis2+x-y-2*2*mod
ll X_Y=dis2+a.x()-a.y()-2*2*mod;
// #ifdef LOCAL
// debug(X_Y);
// #endif
ll X=(XY+X_Y)/2;
ll Y=XY-X;
cout<<"! "<<X<<" "<<Y<<endl;
}
/*
AC
https://codeforces.com/contest/2136/submission/336079080
1
1
0 0
? D 1000000000
1
? D 1000000000
1
? R 1000000000
1
? R 1000000000
4000000000
? U 1000000000
1
? U 1000000000
1
? U 1000000000
1
? U 1000000000
4000000000
! 0 0
*/
```
:::
## 心路历程（WA，TLE，MLE……）