結果
問題 | No.64 XORフィボナッチ数列 |
ユーザー | mikam |
提出日時 | 2023-02-02 17:07:55 |
言語 | C++17 (gcc 12.3.0 + boost 1.83.0) |
結果 |
AC
|
実行時間 | 2 ms / 5,000 ms |
コード長 | 3,305 bytes |
コンパイル時間 | 4,250 ms |
コンパイル使用メモリ | 264,072 KB |
実行使用メモリ | 6,944 KB |
最終ジャッジ日時 | 2024-07-02 05:10:44 |
合計ジャッジ時間 | 4,902 ms |
ジャッジサーバーID (参考情報) |
judge1 / judge3 |
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テストケース
テストケース表示入力 | 結果 | 実行時間 実行使用メモリ |
---|---|---|
testcase_00 | AC | 2 ms
6,812 KB |
testcase_01 | AC | 2 ms
6,940 KB |
testcase_02 | AC | 2 ms
6,940 KB |
testcase_03 | AC | 2 ms
6,940 KB |
testcase_04 | AC | 2 ms
6,944 KB |
testcase_05 | AC | 2 ms
6,940 KB |
testcase_06 | AC | 2 ms
6,944 KB |
testcase_07 | AC | 2 ms
6,944 KB |
testcase_08 | AC | 2 ms
6,940 KB |
testcase_09 | AC | 2 ms
6,940 KB |
testcase_10 | AC | 2 ms
6,940 KB |
testcase_11 | AC | 2 ms
6,944 KB |
testcase_12 | AC | 2 ms
6,944 KB |
testcase_13 | AC | 2 ms
6,944 KB |
ソースコード
#include <atcoder/all> using namespace atcoder; #include <bits/stdc++.h> using namespace std; // #include <boost/multiprecision/cpp_int.hpp> #pragma GCC target("avx2") #pragma GCC optimize("O3") #pragma GCC optimize("unroll-loops") #define rep(i, n) for (int i = 0; i < (int)(n); i++) #define rep2(i,a,b) for (int i = (int)(a); i < (int)(b); i++) #define all(v) v.begin(),v.end() #define inc(x,l,r) ((l)<=(x)&&(x)<(r)) #define Unique(x) sort(all(x)), x.erase(unique(all(x)), x.end()) #define pcnt __builtin_popcountll typedef long long ll; #define int ll using ld = long double; using vi = vector<int>; using vs = vector<string>; using P = pair<int,int>; using vp = vector<P>; // using Bint = boost::multiprecision::cpp_int; template<typename T1,typename T2> bool chmax(T1 &a, const T2 b) {if (a < b) {a = b; return true;} else return false; } template<typename T1,typename T2> bool chmin(T1 &a, const T2 b) {if (a > b) {a = b; return true;} else return false; } template<typename T> using priority_queue_greater = priority_queue<T, vector<T>, greater<T>>; template<typename T> ostream &operator<<(ostream &os,const vector<T> &v){rep(i,v.size())os<<v[i]<<(i+1!=v.size()?" ":"");return os;} template<typename T> istream &operator>>(istream& is,vector<T> &v){for(T &in:v)is>>in;return is;} template<typename T1,typename T2> ostream &operator<< (ostream &os, const pair<T1,T2> &p){os << p.first <<" "<<p.second;return os;} ostream &operator<< (ostream &os, const modint1000000007 &m){os << m.val();return os;} istream &operator>> (istream &is, modint1000000007 &m){ll in;is>>in;m=in;return is;} ostream &operator<< (ostream &os, const modint998244353 &m){os << m.val();return os;} istream &operator>> (istream &is, modint998244353 &m){ll in;is>>in;m=in;return is;} template<class... T> void input(T&... a){(cin>> ... >> a);} template<class T> void print(T& a){cout <<a<< '\n';} template<class T,class... Ts> void print(const T&a, const Ts&... b){cout<< a;(cout<<...<<(cout<<' ',b));cout<<'\n';} #define VI(v,n) vi v(n); input(v) #define INT(...) int __VA_ARGS__; input(__VA_ARGS__) #define STR(...) string __VA_ARGS__; input(__VA_ARGS__) #define CHAR(...) char __VA_ARGS__; input(__VA_ARGS__) int sign(int x){return x>0?1:x<0?-1:0;} int ceil(int x,int y){assert(y!=0);if(sign(x)==sign(y))return (x+y-1)/y;return -((-x/y));} int floor(int x,int y){assert(y!=0);if(sign(x)==sign(y))return x/y;if(y<0)x*=-1,y*=-1;return x/y-(x%y<0);} int abs(int x,int y){return abs(x-y);} bool ins(string s,string t){return s.find(t)!=string::npos;} P operator+ (const P &p, const P &q){ return P{p.first+q.first,p.second+q.second};} P operator- (const P &p, const P &q){ return P{p.first-q.first,p.second-q.second};} void yesno(bool ok,string y="Yes",string n="No"){ cout<<(ok?y:n)<<endl;} void YESNO(bool ok,string y="YES",string n="NO"){ cout<<(ok?y:n)<<endl;} int di[]={-1,0,1,0,-1,-1,1,1}; int dj[]={0,1,0,-1,-1,1,-1,1}; const ll INF = 8e18; //using mint = modint1000000007; //using mint = modint998244353; //mint stom(const string &s,int b=10){mint res = 0;for(auto c:s)res *= b,res += c-'0';return res;} signed main() { cin.tie(0); ios_base::sync_with_stdio(false); cout << fixed << setprecision(20); vi a(3); rep(i,2)cin>>a[i]; INT(n); a[2] = a[0]^a[1]; print(a[n%3]); return 0; }