結果
問題 | No.2503 Typical Path Counting Problem on a Grid |
ユーザー | woodywoody |
提出日時 | 2023-10-16 17:47:03 |
言語 | C++14 (gcc 12.3.0 + boost 1.83.0) |
結果 |
AC
|
実行時間 | 702 ms / 2,000 ms |
コード長 | 6,250 bytes |
コンパイル時間 | 4,443 ms |
コンパイル使用メモリ | 246,772 KB |
実行使用メモリ | 42,496 KB |
最終ジャッジ日時 | 2024-09-16 22:15:33 |
合計ジャッジ時間 | 11,457 ms |
ジャッジサーバーID (参考情報) |
judge5 / judge1 |
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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 | 133 ms
42,240 KB |
testcase_01 | AC | 566 ms
42,368 KB |
testcase_02 | AC | 169 ms
42,240 KB |
testcase_03 | AC | 506 ms
42,368 KB |
testcase_04 | AC | 648 ms
42,496 KB |
testcase_05 | AC | 594 ms
42,368 KB |
testcase_06 | AC | 698 ms
42,368 KB |
testcase_07 | AC | 702 ms
42,240 KB |
testcase_08 | AC | 690 ms
42,368 KB |
testcase_09 | AC | 647 ms
42,368 KB |
ソースコード
#include<bits/stdc++.h> #include<atcoder/all> #define rep(i,b) for(int i=0;i<b;i++) #define rrep(i,b) for(int i=b-1;i>=0;i--) #define rep1(i,b) for(int i=1;i<b;i++) #define repx(i,x,b) for(int i=x;i<b;i++) #define rrepx(i,x,b) for(int i=b-1;i>=x;i--) #define fore(i,a) for(auto& i:a) #define rng(x) (x).begin(), (x).end() #define rrng(x) (x).rbegin(), (x).rend() #define sz(x) ((int)(x).size()) #define pb push_back #define fi first #define se second #define pcnt __builtin_popcountll using namespace std; using namespace atcoder; using ll = long long; using ld = long double; template<typename T> using mpq = priority_queue<T, vector<T>, greater<T>>; template<typename T> bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; } template<typename T> bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; } template<typename T> ll sumv(const vector<T>&a){ll res(0);for(auto&&x:a)res+=x;return res;} bool yn(bool a) { if(a) {cout << "Yes" << endl; return true;} else {cout << "No" << endl; return false;}} #define retval(x) {cout << #x << endl; return;} #define cout2(x,y) cout << x << " " << y << endl; #define coutp(p) cout << p.fi << " " << p.se << endl; #define out cout << ans << endl; #define outd cout << fixed << setprecision(20) << ans << endl; #define outm cout << ans.val() << endl; #define outv fore(yans , ans) cout << yans << "\n"; #define outdv fore(yans , ans) cout << yans.val() << "\n"; #define assertmle(x) if (!(x)) {vi v(3e8);} #define asserttle(x) if (!(x)) {while(1){}} #define coutv(v) {fore(vy , v) {cout << vy << " ";} cout << endl;} #define coutv2(v) fore(vy , v) cout << vy << "\n"; #define coutvm(v) {fore(vy , v) {cout << vy.val() << " ";} cout << endl;} #define coutvm2(v) fore(vy , v) cout << vy.val() << "\n"; using pll = pair<ll,ll>;using pil = pair<int,ll>;using pli = pair<ll,int>;using pii = pair<int,int>;using pdd = pair<ld,ld>; using vi = vector<int>;using vd = vector<ld>;using vl = vector<ll>;using vs = vector<string>;using vb = vector<bool>; using vpii = vector<pii>;using vpli = vector<pli>;using vpll = vector<pll>;using vpil = vector<pil>; using vvi = vector<vector<int>>;using vvl = vector<vector<ll>>;using vvs = vector<vector<string>>;using vvb = vector<vector<bool>>; using vvpii = vector<vector<pii>>;using vvpli = vector<vector<pli>>;using vvpll = vector<vpll>;using vvpil = vector<vpil>; using mint = modint998244353; //using mint = modint1000000007; //using mint = dynamic_modint<0>; using vm = vector<mint>; using vvm = vector<vector<mint>>; vector<int> dx={1,0,-1,0,1,1,-1,-1},dy={0,1,0,-1,1,-1,1,-1}; ll gcd(ll a, ll b) { return a?gcd(b%a,a):b;} ll lcm(ll a, ll b) { return a/gcd(a,b)*b;} #define yes {cout <<"Yes"<<endl;} #define no {cout <<"No"<<endl;} const double eps = 1e-10; const ll LINF = 1001002003004005006ll; const int INF = 1001001001; #ifdef MY_LOCAL_DEBUG #include "./debug/localDebug.cpp" #define showp(p) cerr<<#p<<" = "<<p.fi<<" : "<<p.se<<endl #define show1(a) cerr<<#a<<" = "<<a<<endl #define show2(a,b) cerr<<#a<<" = "<<a<<" : "<<#b<<" = "<<b<<endl #define show3(a,b,c) cerr<<#a<<" = "<<a<<" : "<<#b<<" = "<<b<<" : "<<#c<<" = "<<c<<endl #define show4(a,b,c,d) cerr<<#a<<" = "<<a<<" : "<<#b<<" = "<<b<<" : "<<#c<<" = "<<c<<" : "<<#d<<" = "<<d<<endl #define show5(a,b,c,d,e) cerr<<#a<<" = "<<a<<" : "<<#b<<" = "<<b<<" : "<<#c<<" = "<<c<<" : "<<#d<<" = "<<d<<" : "<<#e<<" = "<<e<<endl #define DEBUG_LINE cout << "DEBUG_LINE : " << __LINE__ << endl #define showv(v) {cout<<#v<<" : "; fore(vy , v) {cout << vy << " ";} cout << endl;} #define showv2(v) {cout<<#v<<endl; fore(vy , v) cout << vy << "\n";} #define showvm(v) {cout<<#v<<" : "; fore(vy , v) {cout << vy.val() << " ";} cout << endl;} #define showvm2(v) {cout<<#v<<endl; fore(vy , v) cout << vy.val() << "\n";} #define showmat(v) {cout<<#v<<endl; fore(row , v) { fore(seg , row) cout << seg << " "; cout << endl;}} #define showmatm(v) {cout<<#v<<endl; fore(row , v) { fore(seg , row) cout << seg.val() << " "; cout << endl;}} #else #define showp(p) #define show1(a) #define show2(a,b) #define show3(a,b,c) #define show4(a,b,c,d) #define show5(a,b,c,d,e) #define DEBUG_LINE #define showv(v) #define showv2(v) #define showvm(v) #define showvm2(v) #define showmat(v) #define showmatm(v) #endif #define overload5(a,b,c,d,e,f, ...) f #define show(...) overload5(__VA_ARGS__, show5, show4, show3, show2, show1)(__VA_ARGS__) vm dp; struct matexp{ int n; const int logmax = 64; vector<vvm> exp; matexp(int n) : n(n){ exp.resize(logmax,vvm(n,vm(n))); } vvm mat_product(vvm& a,vvm& b){ int n1 = sz(a); int m1 = sz(a[0]); int n2 = sz(b); int m2 = sz(b[0]); assert(m1 == n2); vvm ret(n1,vm(m2)); rep(i,n1) rep(j,m2) rep(k,m1) ret[i][j] += a[i][k]*b[k][j]; return ret; } void init(vvm a){ exp[0] = a; rep1(i,logmax) exp[i] = mat_product(exp[i-1],exp[i-1]); } vvm getexp_k(ll k){ vvm ret(n,vm(n)); rep(i,n) ret[i][i] = 1; rep(i,logmax) if ((k >> i & 1) == 1) ret = mat_product(ret,exp[i]); return ret; } }; // A^k (Aは行列)を計算 // matexp(n) : n(行列の次元)を初期化 // mat_product(a,b) : 行列積a*bを返す // init(a) : 行列Aを初期化 // getexp_k : A^kを返す void solve(){ ll n,m; cin>>n>>m; if (n > m) swap(n,m); if (n==0) retval(1); mint ans = 0; if (n==m){ ans += dp[n]*dp[n]; ans += dp[n-1]*dp[n-1]*n; }else{ matexp mt(2); vvm a(2,vm(2)); a[0][0] = n*2 + 1; a[0][1] = n; a[1][0] = 1; a[1][1] = 0; mt.init(a); a = mt.getexp_k(m-n-1); mint fir = dp[n-1]*n + dp[n]*(2*n+1); mint sec = dp[n]; mint fir2 = a[0][0]*fir + a[0][1]*sec; mint sec2 = a[1][0]*fir + a[1][1]*sec; ans += fir2 * sec; ans += sec2 * dp[n-1]*n; } outm; return; } int main(){ ios::sync_with_stdio(false); cin.tie(0); int t = 1; cin>>t; const int nmax = (int)1e7 + 1; dp.resize(nmax); dp[0] = 1; dp[1] = 2; repx(i,2,nmax){ dp[i] = dp[i-1]*2*i + dp[i-2]*(i-1); } rep(i,t){ solve(); } return 0; }