結果
問題 | No.471 直列回転機 |
ユーザー |
![]() |
提出日時 | 2016-12-21 00:14:20 |
言語 | C++14 (gcc 13.3.0 + boost 1.87.0) |
結果 |
TLE
(最新)
AC
(最初)
|
実行時間 | - |
コード長 | 4,282 bytes |
コンパイル時間 | 1,847 ms |
コンパイル使用メモリ | 175,332 KB |
実行使用メモリ | 40,392 KB |
最終ジャッジ日時 | 2024-07-16 11:31:15 |
合計ジャッジ時間 | 11,243 ms |
ジャッジサーバーID (参考情報) |
judge4 / judge5 |
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ファイルパターン | 結果 |
---|---|
sample | TLE * 1 |
other | -- * 58 |
ソースコード
// #includes {{{#include <bits/stdc++.h>using namespace std;// }}}// pre-written code {{{#define REP(i,n) for(int i=0;i<(int)(n);++i)#define RREP(i,a,b) for(int i=(int)(a);i<(int)(b);++i)#define FOR(i,c) for(__typeof((c).begin()) i=(c).begin();i!=(c).end();++i)#define LET(x,a) __typeof(a) x(a)//#define IFOR(i,it,c) for(__typeof((c).begin())it=(c).begin();it!=(c).end();++it,++i)#define ALL(c) (c).begin(), (c).end()#define MP make_pair#define EXIST(e,s) ((s).find(e)!=(s).end())#define RESET(a) memset((a),0,sizeof(a))#define SET(a) memset((a),-1,sizeof(a))#define PB push_back#define DEC(it,command) __typeof(command) it=command//debug#define dump(x) cerr << #x << " = " << (x) << endl;#define debug(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl;#define debug2(x) cerr << #x << " = [";REP(__ind,(x).size()){cerr << (x)[__ind] << ", ";}cerr << "] (L" << __LINE__ << ")" << endl;const int INF=0x3f3f3f3f;typedef long long Int;typedef unsigned long long uInt;#ifdef __MINGW32__typedef double rn;#elsetypedef long double rn;#endiftypedef pair<int,int> pii;/*#ifdef MYDEBUG#include"debug.h"#include"print.h"#endif*/// }}}typedef rn Num;const Num mod = 998244353;//{{{ mattypedef rn nm;typedef vector<nm> vec;typedef vector<vec> mat;//int/*nm add(const nm &x,const nm &y){return (x+y)%mod;}nm opposite(const nm &n){return ((-n)%mod+mod)%mod;}nm mul(const nm &x,const nm &y){return (unsigned long long)x*y%mod;}//nm inverse(const nm &n){return invMod(n,mod);}//nm modulo(const nm &n,int mod=::mod){return ((n+mod)%mod+mod)%mod;}*///real numbernm add(const nm &x,const nm &y){return x+y;}nm opposite(const nm &n){return -n;}nm mul(const nm &x,const nm &y){return x*y;}nm inverse(const nm &n){return 1L/n;}// O( n )mat identity(int n) {mat A(n, vec(n));for (int i = 0; i < n; ++i) A[i][i] = 1;return A;}// O( n )nm inner_product(const vec &a, const vec &b) {nm ans = 0;for (int i = 0; i < a.size(); ++i)ans = add(ans,mul(a[i],b[i]));return ans;}mat add(const mat &A, const mat &B) {mat C(A.size(), vec(B[0].size()));for (int i = 0; i < C.size(); ++i)for (int j = 0; j < C[i].size(); ++j)C[i][j] = add(A[i][j],B[i][j]);return C;}// O( n^2 )vec mul(const mat &A, const vec &x) {vec y(A.size());for (int i = 0; i < A.size(); ++i)for (int j = 0; j < A[0].size(); ++j)y[i] = add(y[i], mul(A[i][j],x[j]));return y;}// O( n^3 )mat mul(const mat &A, const mat &B) {mat C(A.size(), vec(B[0].size()));for (int i = 0; i < C.size(); ++i)for (int j = 0; j < C[i].size(); ++j)for (int k = 0; k < A[i].size(); ++k)C[i][j] = add(C[i][j],mul(A[i][k],B[k][j]));return C;}// O( n^3 log e )mat pow(const mat &A, long long e) {return e == 0 ? identity(A.size()) :e % 2 == 0 ? pow(mul(A, A), e/2) : mul(A, pow(A, e-1));}//}}}//{{{ gaussint gauss(mat &A, vec &b) {//int gauss(mat& A) {//returns rankconst int n=A.size(),m=A[0].size();int pi=0;for(int pj=0;pj<m;pj++){for(int i = pi+1; i < n; i++) {if (abs(A[i][pj]) > abs(A[pi][pj])) {swap(A[i], A[pi]);swap(b[i], b[pi]);}}if (abs(A[pi][pj]) > 0) {nm d = inverse(A[pi][pj]);REP(j, m)A[pi][j] = mul(A[pi][j],d);b[pi] = mul(b[pi],d);REP(i,n){if(i==pi)continue;nm k = A[i][pj];REP(j, m)A[i][j] = add(A[i][j],opposite(mul(k, A[pi][j])));b[i] = add(b[i],opposite(mul(k, b[pi])));}pi++;}}return pi;/*for(int i = pi; i < n; i++)if (abs(b[i]) > 0)throw Inconsistent();if (pi < m || pj < m)throw Ambiguous();for(int j = m-1; j >= 0; j--)REP(i, j)b[i] = modulo(b[i] - b[j] * A[i][j]);*/}//}}}int main(){const int M = 3;mat A(6,vec(6));vec b(6);REP(i,M){int x = rand()%20000-10000;int y = rand()%20000-10000;printf("? %d %d\n",x,y);long long a0,b0;scanf("%lld%lld",&a0,&b0);A[i*2][0] = (rn)x;A[i*2][1] = (rn)y;A[i*2][2] = 1.0L;A[i*2+1][0] = (rn)x;A[i*2+1][1] = (rn)y;A[i*2+1][2] = 1.0L;b[i*2] = (rn)a0;b[i*2+1] = (rn)b0;}gauss(A,b);printf("!\n");int N;scanf("%d",&N);REP(i,N){rn x,y;scanf("%Lf%Lf",&x,&y);printf("%Lf %Lf\n",b[0]*x+b[1]*y+b[2],b[3]*x+b[4]*y+b[5]);}return 0;}