oj.uz

답안 #57730

# 제출 시각 아이디 문제 언어 결과 실행 시간 메모리
57730 2018-07-15T23:56:19 Z Benq Multiply (CEOI17_mul) C++14
100 / 100
 90 ms 6792 KB 
#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>

using namespace std;
using namespace __gnu_pbds;
 
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;

typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;

typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;

template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;

#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)

#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()

const int MOD = 1000000007;
const ll INF = 1e18;
const int MX = 100001;

namespace NTT {
    const ll mod = (119 << 23) + 1, root = 3; // = 998244353
    // For p < 2^30 there is also e.g. (5 << 25, 3), (7 << 26, 3),
    // (479 << 21, 3) and (483 << 21, 5). The last two are > 10^9.
        
    ll modpow(ll b, ll p) { return !p?1:modpow(b*b%mod,p/2)*(p&1?b:1)%mod; }
    
    ll inv (ll b) { return modpow(b,mod-2); }
    
    int get(int s) {
        return s > 1 ? 32 - __builtin_clz(s - 1) : 0;
    }
    
    vl ntt(vl& a) { 
        int n = a.size(), x = get(n); 
        vl res, RES(n), roots(n);
        roots[0] = 1, roots[1] = modpow(root,(mod-1)/n);
        FOR(i,2,n) roots[i] = roots[i-1]*roots[1] % mod;
        
        res = a;
        FOR(i,1,x+1) {
            int inc = n>>i;
            F0R(j,inc) for (int k = 0; k < n; k += inc) {
                int t = 2*k%n+j;
                RES[k+j] = (res[t]+roots[k]*res[t+inc]) % mod;
            }
            swap(res,RES);
        }
        
        return res;
    }
    
    vl ntt_rev(vl& a) {
        vl res = ntt(a);
        ll in = inv(a.size());
        F0R(i,sz(res)) res[i] = res[i]*in % mod;
        reverse(res.begin() + 1, res.end());
        return res;
    }
    
    vl brute(vl& a, vl& b) {
        vl c(sz(a)+sz(b)-1);
        F0R(i,sz(a)) F0R(j,sz(b)) c[i+j] = (c[i+j]+a[i]*b[j])%mod;
        return c;
    }
    
    vl conv(vl a, vl b) {
        int s = sz(a)+sz(b)-1, L = get(s), n = 1<<L;
        if (s <= 0) return {};
        if (s <= 200) return brute(a,b);
        
        a.resize(n); a = ntt(a);
        b.resize(n); b = ntt(b);
        
        F0R(i,n) a[i] = a[i]*b[i] % mod;
        a = ntt_rev(a);
        
        a.resize(s);
        return a;
    }
}

int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    int a,b; cin >> a >> b;
    vl A(a), B(b);
    F0R(i,a) {
        char c; cin >> c;
        A[a-1-i] = c-'0';
    }
    F0R(i,b) {
        char c; cin >> c;
        B[b-1-i] = c-'0';
    }
    vl C = NTT::conv(A,B);
    F0R(i,sz(C)) if (C[i] >= 10) {
        if (i+1 == sz(C)) C.pb(0);
        C[i+1] += C[i]/10; C[i] %= 10;
    }
    while (sz(C) > 1 && C.back() == 0) C.pop_back();
    F0Rd(i,sz(C)) cout << C[i];
}

/* Look for:
* the exact constraints (multiple sets are too slow for n=10^6 :( ) 
* special cases (n=1?)
* overflow (ll vs int?)
* array bounds
*/

Subtask #1
20.0 / 20.0

# 결과 실행 시간 메모리 Grader output
1 Correct 2 ms 376 KB Output is correct
2 Correct 2 ms 496 KB Output is correct
3 Correct 2 ms 496 KB Output is correct
4 Correct 2 ms 496 KB Output is correct
5 Correct 3 ms 572 KB Output is correct

Subtask #2
20.0 / 20.0

# 결과 실행 시간 메모리 Grader output
1 Correct 2 ms 376 KB Output is correct
2 Correct 2 ms 496 KB Output is correct
3 Correct 2 ms 496 KB Output is correct
4 Correct 2 ms 496 KB Output is correct
5 Correct 3 ms 572 KB Output is correct
6 Correct 2 ms 620 KB Output is correct
7 Correct 2 ms 684 KB Output is correct
8 Correct 2 ms 688 KB Output is correct
9 Correct 2 ms 692 KB Output is correct
10 Correct 2 ms 716 KB Output is correct

Subtask #3
30.0 / 30.0

# 결과 실행 시간 메모리 Grader output
1 Correct 2 ms 376 KB Output is correct
2 Correct 2 ms 496 KB Output is correct
3 Correct 2 ms 496 KB Output is correct
4 Correct 2 ms 496 KB Output is correct
5 Correct 3 ms 572 KB Output is correct
6 Correct 2 ms 620 KB Output is correct
7 Correct 2 ms 684 KB Output is correct
8 Correct 2 ms 688 KB Output is correct
9 Correct 2 ms 692 KB Output is correct
10 Correct 2 ms 716 KB Output is correct
11 Correct 5 ms 1100 KB Output is correct
12 Correct 5 ms 1148 KB Output is correct
13 Correct 5 ms 1176 KB Output is correct
14 Correct 10 ms 1552 KB Output is correct
15 Correct 3 ms 1552 KB Output is correct

Subtask #4
30.0 / 30.0

# 결과 실행 시간 메모리 Grader output
1 Correct 2 ms 376 KB Output is correct
2 Correct 2 ms 496 KB Output is correct
3 Correct 2 ms 496 KB Output is correct
4 Correct 2 ms 496 KB Output is correct
5 Correct 3 ms 572 KB Output is correct
6 Correct 2 ms 620 KB Output is correct
7 Correct 2 ms 684 KB Output is correct
8 Correct 2 ms 688 KB Output is correct
9 Correct 2 ms 692 KB Output is correct
10 Correct 2 ms 716 KB Output is correct
11 Correct 5 ms 1100 KB Output is correct
12 Correct 5 ms 1148 KB Output is correct
13 Correct 5 ms 1176 KB Output is correct
14 Correct 10 ms 1552 KB Output is correct
15 Correct 3 ms 1552 KB Output is correct
16 Correct 45 ms 3744 KB Output is correct
17 Correct 90 ms 6732 KB Output is correct
18 Correct 43 ms 6732 KB Output is correct
19 Correct 87 ms 6792 KB Output is correct
20 Correct 39 ms 6792 KB Output is correct