結果
問題 | No.1084 積の積 |
ユーザー | Thistle |
提出日時 | 2020-06-20 01:51:23 |
言語 | C++14 (gcc 12.3.0 + boost 1.83.0) |
結果 |
WA
|
実行時間 | - |
コード長 | 6,286 bytes |
コンパイル時間 | 1,794 ms |
コンパイル使用メモリ | 132,692 KB |
実行使用メモリ | 13,368 KB |
最終ジャッジ日時 | 2024-07-03 16:34:15 |
合計ジャッジ時間 | 2,872 ms |
ジャッジサーバーID (参考情報) |
judge5 / judge2 |
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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 | WA | - |
testcase_01 | WA | - |
testcase_02 | WA | - |
testcase_03 | WA | - |
testcase_04 | WA | - |
testcase_05 | WA | - |
testcase_06 | AC | 6 ms
12,392 KB |
testcase_07 | WA | - |
testcase_08 | AC | 5 ms
11,732 KB |
testcase_09 | WA | - |
testcase_10 | WA | - |
testcase_11 | WA | - |
testcase_12 | WA | - |
testcase_13 | WA | - |
testcase_14 | WA | - |
testcase_15 | WA | - |
testcase_16 | WA | - |
testcase_17 | WA | - |
testcase_18 | WA | - |
testcase_19 | WA | - |
testcase_20 | WA | - |
testcase_21 | WA | - |
testcase_22 | WA | - |
testcase_23 | WA | - |
testcase_24 | WA | - |
testcase_25 | WA | - |
testcase_26 | WA | - |
testcase_27 | WA | - |
testcase_28 | WA | - |
testcase_29 | WA | - |
testcase_30 | WA | - |
testcase_31 | WA | - |
コンパイルメッセージ
main.cpp: In function 'int main()': main.cpp:180:29: warning: 'mul' may be used uninitialized [-Wmaybe-uninitialized] 180 | while (r < n && mul * a[r] < 1000000000) { main.cpp:169:16: note: 'mul' was declared here 169 | int r = 0, mul; | ^~~
ソースコード
#pragma GCC target ("avx") #pragma GCC optimize("Ofast") #pragma GCC optimize("unroll-loops") //#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native") #define _USE_MATH_DEFINES #include<iostream> #include<string> #include<queue> #include<cmath> #include<map> #include<set> #include<list> #include<iomanip> #include<vector> #include<random> #include<functional> #include<algorithm> #include<stack> #include<cstdio> #include<cstring> #include<bitset> #include<unordered_map> #include<climits> #include<fstream> #include<complex> #include<time.h> #include<cassert> #include<functional> #include<numeric> #include<tuple> using namespace std; using ll = long long; using ld = long double; #define int long long #define all(a) (a).begin(),(a).end() #define fs first #define sc second #define xx first #define yy second.first #define zz second.second #define H pair<int, int> #define P pair<int, pair<int, int>> #define Q(i,j,k) mkp(i,mkp(j,k)) #define rng(i,s,n) for(int i = (s) ; i < (n) ; i++) #define rep(i,n) rng(i, 0, (n)) #define mkp make_pair #define vec vector #define vi vec<int> #define pb emplace_back #define siz(a) (int)(a).size() #define crdcomp(b) sort(all((b)));(b).erase(unique(all((b))),(b).end()) #define getidx(b,i) lower_bound(all(b),(i))-b.begin() #define ssp(i,n) (i==(int)(n)-1?"\n":" ") #define ctoi(c) (int)(c-'0') #define itoc(c) (char)(c+'0') #define cyes printf("Yes\n") #define cno printf("No\n") #define cdf(n) int quetimes_=(n);rep(qq123_,quetimes_) #define gcj printf("Case #%lld: ",qq123_+1) #define readv(a,n) a.resize(n,0);rep(i,(n)) a[i]=read() //#define endl "\n" constexpr int mod = 1e9 + 7; constexpr int Mod = 998244353; constexpr ld EPS = 1e-10; constexpr ll inf = 3 * 1e18; constexpr int Inf = 15 * 1e8; constexpr int dx[] = { -1,1,0,0 }, dy[] = { 0,0,-1,1 }; template<class T>bool chmax(T& a, const T& b) { if (a < b) { a = b; return 1; } return 0; } template<class T>bool chmin(T& a, const T& b) { if (b < a) { a = b; return 1; } return 0; } ll read() { ll u, k = scanf("%lld", &u); return u; } string reads() { string s; cin >> s; return s; } H readh(bool g = 0) { H u; int k = scanf("%lld %lld", &u.fs, &u.sc); if (g) u.fs--, u.sc--; return u; } bool ina(H t, int h, int w) { return 0 <= t.fs && t.fs < h && 0 <= t.sc && t.sc < w; } bool ina(int t, int l, int r) { return l <= t && t < r; } ll gcd(ll i, ll j) { return j ? gcd(j, i % j) : i; } ll popcount(ll x) { int sum = 0; for (int i = 0; i < 60; i++)if ((1ll << i) & x) sum++; return sum; } class mint { public:ll v; mint(ll v = 0) { s(v % mod + mod); } constexpr static int mod = 1e9 + 7; constexpr static int fn_ = 500005; static mint fact[fn_], comp[fn_]; mint pow(mint x) const { mint b(v), c(1); while (x.v) { if (x.v & 1) c *= b; b *= b; x.v >>= 1; } return c; } inline mint& s(int vv) { v = vv < mod ? vv : vv - mod; return *this; } inline mint inv()const { return pow(mod - 2); } inline mint operator-()const { return mint() - *this; } inline mint& operator+=(const mint b) { return s(v + b.v); } inline mint& operator-=(const mint b) { return s(v + mod - b.v); } inline mint& operator*=(const mint b) { v = v * b.v % mod; return *this; } inline mint& operator/=(const mint b) { v = v * b.inv().v % mod; return *this; } inline mint operator+(const mint b) const { return mint(v) += b; } inline mint operator-(const mint b) const { return mint(v) -= b; } inline mint operator*(const mint b) const { return mint(v) *= b; } inline mint operator/(const mint b) const { return mint(v) /= b; } friend ostream& operator<<(ostream& os, const mint& m) { return os << m.v; } friend istream& operator>>(istream& is, mint& m) { int x; is >> x; m = mint(x); return is; } bool operator<(const mint& r)const { return v < r.v; } bool operator>(const mint& r)const { return v > r.v; } bool operator<=(const mint& r)const { return v <= r.v; } bool operator>=(const mint& r)const { return v >= r.v; } bool operator==(const mint& r)const { return v == r.v; } bool operator!=(const mint& r)const { return v != r.v; } explicit operator bool()const { return v; } explicit operator int()const { return v; } mint comb(mint k) { if (k > * this) return mint(); if (!fact[0]) combinit(); if (v >= fn_) { if (k > * this - k) k = *this - k; mint tmp(1); for (int i = v; i >= v - k.v + 1; i--) tmp *= mint(i); return tmp * comp[k.v]; } else return fact[v] * comp[k.v] * comp[v - k.v]; }//nCk static void combinit() { fact[0] = 1; for (int i = 1; i < fn_; i++) fact[i] = fact[i - 1] * mint(i); comp[fn_ - 1] = fact[fn_ - 1].inv(); for (int i = fn_ - 2; i >= 0; i--) comp[i] = comp[i + 1] * mint(i + 1); } }; mint mint::fact[fn_], mint::comp[fn_]; //-------------------------------------------------------------- //--------------------------------------------------------------------- int b[500000]; //現実の総和 int c[500000]; //この時点での階段が何個あるか? int mod_pow(int x, int v) { int ret = 1; while (v > 0) { if (v & 1) (ret *= x) %= mod; (x *= x) %= mod; v >>= 1; } return ret; } signed main() { freopen("C:/Users/Thistle/Downloads/3681-testcase/test_in/1_corner_1.txt", "r", stdin); int n; cin >> n; vec<int>a; readv(a, n); int r = 0, mul; if (count(all(a), 0)) { cout << 0 << endl; return 0; } rep(l, n) { if (l == r) { r = l + 1; mul = a[l]; } while (r < n && mul * a[r] < 1000000000) { mul *= a[r]; r++; } b[l] += r - l; c[l + 1] += 1; c[r + 1] -= 1; mul /= a[l]; } int ans = 1; rep(i, n) { if (i > 0) b[i] += b[i - 1], c[i] += c[i - 1]; b[i] -= c[i]; ans *= mod_pow(a[i], b[i]); ans %= mod; } cout << ans << endl; }