結果
問題 | No.2156 ぞい文字列 |
ユーザー | 👑 deuteridayo |
提出日時 | 2022-12-09 22:10:20 |
言語 | C++17 (gcc 13.2.0 + boost 1.83.0) |
結果 |
AC
|
実行時間 | 2 ms / 2,000 ms |
コード長 | 3,438 bytes |
コンパイル時間 | 4,472 ms |
コンパイル使用メモリ | 269,720 KB |
実行使用メモリ | 4,380 KB |
最終ジャッジ日時 | 2023-08-04 23:53:02 |
合計ジャッジ時間 | 5,527 ms |
ジャッジサーバーID (参考情報) |
judge15 / judge12 |
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 | 1 ms
4,380 KB |
testcase_01 | AC | 2 ms
4,376 KB |
testcase_02 | AC | 1 ms
4,376 KB |
testcase_03 | AC | 2 ms
4,376 KB |
testcase_04 | AC | 2 ms
4,380 KB |
testcase_05 | AC | 1 ms
4,380 KB |
testcase_06 | AC | 1 ms
4,376 KB |
testcase_07 | AC | 1 ms
4,380 KB |
testcase_08 | AC | 2 ms
4,376 KB |
testcase_09 | AC | 2 ms
4,380 KB |
testcase_10 | AC | 2 ms
4,380 KB |
testcase_11 | AC | 2 ms
4,376 KB |
testcase_12 | AC | 1 ms
4,376 KB |
testcase_13 | AC | 1 ms
4,376 KB |
testcase_14 | AC | 1 ms
4,380 KB |
testcase_15 | AC | 1 ms
4,380 KB |
testcase_16 | AC | 2 ms
4,376 KB |
testcase_17 | AC | 1 ms
4,380 KB |
testcase_18 | AC | 1 ms
4,380 KB |
testcase_19 | AC | 2 ms
4,376 KB |
ソースコード
#include<bits/stdc++.h> #include<atcoder/all> using namespace std; using namespace atcoder; using lint = long long; using ulint = unsigned long long; #define endl '\n' int const INF = 1<<30; lint const INF64 = 1LL<<61; // lint const mod = 1e9+7; // using mint = modint1000000007; long const mod = 998244353; using mint = modint998244353; lint ceilDiv(lint x, lint y){if(x >= 0){return (x+y-1)/y;}else{return x/y;}} lint floorDiv(lint x, lint y){if(x >= 0){return x/y;}else{return (x-y+1)/y;}} lint Sqrt(lint x){lint upper = 1e9;lint lower = 0;while(upper - lower > 0){lint mid = (1+upper + lower)/2;if(mid * mid > x){upper = mid-1;}else{lower = mid;}}return upper;} lint gcd(lint a,lint b){if(a<b)swap(a,b);if(a%b==0)return b;else return gcd(b,a%b);} lint lcm(lint a,lint b){return (a / gcd(a,b)) * b;} lint chmin(vector<lint>&v){lint ans = INF64;for(lint i:v){ans = min(ans, i);}return ans;} lint chmax(vector<lint>&v){lint ans = -INF64;for(lint i:v){ans = max(ans, i);}return ans;} double dist(double x1, double y1, double x2, double y2){return sqrt(pow(x1-x2, 2) + pow(y1-y2,2));} string toString(lint n){string ans = "";if(n == 0){ans += "0";}else{while(n > 0){int a = n%10;char b = '0' + a;string c = "";c += b;n /= 10;ans = c + ans;}}return ans;} string toString(lint n, lint k){string ans = toString(n);string tmp = "";while(ans.length() + tmp.length() < k){tmp += "0";}return tmp + ans;} vector<lint>prime;void makePrime(lint n){prime.push_back(2);for(lint i=3;i<=n;i+=2){bool chk = true;for(lint j=0;j<prime.size() && prime[j]*prime[j] <= i;j++){if(i % prime[j]==0){chk=false;break;}}if(chk)prime.push_back(i);}} lint Kai[20000001]; bool firstCallnCr = true; lint ncrmodp(lint n,lint r,lint p){ if(firstCallnCr){ Kai[0] = 1; for(int i=1;i<=20000000;i++){ Kai[i] = Kai[i-1] * i; Kai[i] %= mod;} firstCallnCr = false;} if(n<0)return 0; if(n < r)return 0;if(n==0)return 1;lint ans = Kai[n];lint tmp = (Kai[r] * Kai[n-r]) % p;for(lint i=1;i<=p-2;i*=2){if(i & p-2){ans *= tmp;ans %= p;}tmp *= tmp;tmp %= p;}return ans;} #define rep(i, n) for(int i = 0; i < n; i++) #define repp(i, x, y) for(int i = x; i < y; i++) #define vec vector #define pb push_back #define se second #define fi first struct edge{ int to; }; using graph = vector<vector<edge>>; int main(){ lint n; cin >> n; vec<vec<vec<mint>>>M(100, vec<vec<mint>>(2, vec<mint>(2, 0))); M[0][0][0] = 1; M[0][0][1] = 1; M[0][1][0] = 1; M[0][1][1] = 0; for(int i = 0; i < 80; i++) { mint a = M[i][0][0]; mint b = M[i][1][0]; mint c = M[i][1][1]; M[i+1][0][0] = a*a + b*b; M[i+1][0][1] = a*b + b*c; M[i+1][1][0] = a*b + b*c; M[i+1][1][1] = b*b + c*c; } // cout << M[2][0][0].val() << " " << M[2][0][1].val() << endl << M[2][1][0].val() << " " << M[2][1][1].val(); vec<vec<mint>>ans(2, vec<mint>(2)); ans[0][0] = 1; ans[1][1] = 1; rep(i, 63) { if((1LL << i) & n) { mint a = M[i][0][0]; mint b = M[i][0][1]; mint c = M[i][1][0]; mint d = M[i][1][1]; mint A = ans[0][0]; mint B = ans[0][1]; mint C = ans[1][0]; mint D = ans[1][1]; ans[0][0] = A*a + B*c; ans[0][1] = A*b + B*d; ans[1][0] = C*a + D*c; ans[1][1] = C*b + D*d; } } cout << (ans[0][0] + ans[1][0] - 1 - ans[0][1]).val() << endl; }