結果
問題 | No.1253 雀見椪 |
ユーザー | kenta255 |
提出日時 | 2020-10-10 09:36:03 |
言語 | C++17 (gcc 12.3.0 + boost 1.83.0) |
結果 |
AC
|
実行時間 | 163 ms / 2,000 ms |
コード長 | 3,244 bytes |
コンパイル時間 | 3,045 ms |
コンパイル使用メモリ | 209,272 KB |
実行使用メモリ | 5,376 KB |
最終ジャッジ日時 | 2024-07-20 15:40:28 |
合計ジャッジ時間 | 4,473 ms |
ジャッジサーバーID (参考情報) |
judge4 / 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
5,248 KB |
testcase_01 | AC | 2 ms
5,376 KB |
testcase_02 | AC | 3 ms
5,376 KB |
testcase_03 | AC | 12 ms
5,376 KB |
testcase_04 | AC | 149 ms
5,376 KB |
testcase_05 | AC | 151 ms
5,376 KB |
testcase_06 | AC | 151 ms
5,376 KB |
testcase_07 | AC | 150 ms
5,376 KB |
testcase_08 | AC | 150 ms
5,376 KB |
testcase_09 | AC | 150 ms
5,376 KB |
testcase_10 | AC | 163 ms
5,376 KB |
testcase_11 | AC | 151 ms
5,376 KB |
testcase_12 | AC | 152 ms
5,376 KB |
testcase_13 | AC | 150 ms
5,376 KB |
testcase_14 | AC | 1 ms
5,376 KB |
ソースコード
#include <bits/stdc++.h> using namespace std; typedef long long ll; #define P pair<ll,ll> #define FOR(I,A,B) for(ll I = ll(A); I < ll(B); ++I) #define FORR(I,A,B) for(ll I = ll((B)-1); I >= ll(A); --I) #define TO(x,t,f) ((x)?(t):(f)) #define SORT(x) (sort(x.begin(),x.end())) // 0 2 2 3 4 5 8 9 #define POSL(x,v) (lower_bound(x.begin(),x.end(),v)-x.begin()) //xi>=v x is sorted #define POSU(x,v) (upper_bound(x.begin(),x.end(),v)-x.begin()) //xi>v x is sorted #define NUM(x,v) (POSU(x,v)-POSL(x,v)) //x is sorted #define REV(x) (reverse(x.begin(),x.end())) //reverse ll gcd_(ll a,ll b){if(a%b==0)return b;return gcd_(b,a%b);} ll lcm_(ll a,ll b){ll c=gcd_(a,b);return ((a/c)*b);} #define NEXTP(x) next_permutation(x.begin(),x.end()) const ll INF=ll(1e16)+ll(7); const ll MOD=1000000007LL; #define out(a) cout<<fixed<<setprecision((a)) //tie(a,b,c) = make_tuple(10,9,87); #define pop_(a) __builtin_popcount((a)) ll keta(ll a){ll r=0;while(a){a/=10;r++;}return r;} class comb{ vector<ll> f,fr; ll MOD_; public: //a^(p-1) = 1 (mod p)(p->Prime numbers) //a^(p-2) = a^(-1) ll calc(ll a,ll b,ll p){//a^(b) mod p if(b==0)return 1; ll y = calc(a,b/2,p);y=(y*y)%p; if(b & 1) y = (y * a) % p; return y; } void init(ll n,ll mod){//input max_n MOD_ = mod; f.resize(n+1); fr.resize(n+1); f[0]=fr[0]=1; for(ll i=1;i<n+1;i++){ f[i] = (f[i-1] * i) % mod; } fr[n] = calc(f[n],mod-2,mod); for(ll i=n-1;i>=0;i--){ fr[i] = fr[i+1] * (i+1) % mod; } } ll nCr(ll n,ll r){ if(n<0||r<0||n<r)return 0; return f[n] * fr[r] % MOD_ * fr[n-r] % MOD_; }//nHr = n+r-1Cr }; class modll{ public: ll v,m; modll(ll v=0,ll m=MOD):v(v),m(m){} modll operator + (modll p) {return modll( ( v + p.v ) % m , m );} modll operator - (modll p) {return modll( ( v + p.m - p.v ) % p.m , m );} modll operator * (modll a) {return modll(v * a.v % a.m, a.m);} modll operator / (modll a) {return modll(v * calcm(a.v,a.m-2,a.m) % a.m , m );} modll operator ^ (modll a) {return modll(calcm(v,a.v,a.m),a.m);} modll operator + (ll p) {return modll( ( v + p ) % m , m );} modll operator - (ll p) {return modll( ( v + m - p ) % m , m );} modll operator * (ll p) {return modll( ( v * p ) % m , m );} modll operator / (ll a) {return modll(v*calcm(a,m-2,m)%m,m);} modll operator ^ (ll a) {return modll(calcm(v,a,m),m);} ll calcm(ll a,ll b,ll p){//a^(b) mod p if(b==0)return 1; ll y = calcm(a,b/2,p);y=(y*y)%p; if(b & 1) y = (y * a) % p; return y; } }; ll calc(ll a,ll b,ll p){//a^(b) mod p if(b==0)return 1; ll y = calc(a,b/2,p);y=(y*y)%p; if(b & 1) y = (y * a) % p; return y; } modll mpow(modll a,modll b){ return modll(calc(a.v,b.v,a.m),a.m); } int main(){ ll T; cin >> T; FOR(t,0,T){ modll N; vector<modll> A(3),B(3); cin >> N.v; N.m = MOD; FOR(i,0,3) cin >> A[i].v >> B[i].v; FOR(i,0,3) A[i].m = B[i].m = MOD; modll only0 = (A[0] / B[0]) ^ N; modll only1 = (A[1] / B[1]) ^ N; modll only2 = (A[2] / B[2]) ^ N; modll not0 = (modll(1,MOD) - (A[0] / B[0])) ^ N; modll not1 = (modll(1,MOD) - (A[1] / B[1])) ^ N; modll not2 = (modll(1,MOD) - (A[2] / B[2])) ^ N; modll ans = not0+not1+not2-only1-only1-only0-only0-only2-only2;; ans = modll(1,MOD) - ans; cout << ans.v << endl; } }