Submission #60213

#	Time	Username	Problem	Language	Result	Execution time	Memory
60213		Benq	Bowling (BOI15_bow)	C++11	42 / 100	1074 ms	1016 KiB

bow

#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;

#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;


struct hsh {
    size_t operator()(const pi& k) const {
        return (ll)MOD*k.f+k.s; // bad, but you get the point
    }
};

int q, n, num[10];
unordered_map<pi,ll,hsh> m[2][3], M[2][3];
string tmp[10];

string gen(vi nex) {
    int lst = -1;
    string res;
    for (int i: nex) {
        if (i == -1) res += '-';
        else if (lst != -1 && i+lst == 10) {
            res += '/';
            lst = -1;
        } else if (i == 10) res += 'x';
        else {
            if (lst == -1) lst = 0;
            res += char('0'+i);
            lst += i;
        }
    }
    return res;
}

bool match(string bad, vi nex) {
    string t = gen(nex);
    F0R(i,sz(bad)) if (bad[i] != '?' && bad[i] != t[i]) return 0;
    return 1;
}


void tri(int X, int Y, pair<pi, ll> A, int x, vi y) {
    if (!match(tmp[x],y)) return;
    pi a = A.f;
    
    int cur = a.s;
    for (int i: y) if (i != -1) {
        cur += i;
        if (X) {
            a.f += i, a.s += i, cur += i;
            X --;
        }
        if (Y) {
            a.s += i, cur += i;
            Y --;
        }
    }
    int posi = 0;
    if (x != n-1) {
        if (y[0] == 10) posi = 2;
        else if (y[0]+y[1] == 10) posi = 1;
    }
    if (x >= 2 && X == 0 && num[x-2] != -1 && a.f != num[x-2]) return;
    if (x >= 1 && Y == 0 && num[x-1] != -1 && a.s != num[x-1]) return;
    if (posi == 0 && num[x] != -1 && cur != num[x]) return;
    M[Y][posi][{a.s,cur}] += A.s;
}

void process(int x) {
    if (x != n-1) {
        F0R(X,2) F0R(Y,3) for (auto a: m[X][Y]) {
            tri(X,Y,a,x,{10,-1});
            F0R(i,10) F0R(j,11-i) tri(X,Y,a,x,{i,j});
        }
    } else {
        F0R(X,2) F0R(Y,3) for (auto a: m[X][Y]) {
            F0R(i,11) tri(X,Y,a,x,{10,10,i});
            F0R(i,10) F0R(j,11-i) tri(X,Y,a,x,{10,i,j});
            F0R(i,10) F0R(j,11) tri(X,Y,a,x,{i,10-i,j});
            F0R(i,10) F0R(j,10-i) tri(X,Y,a,x,{i,j,-1});
        }
    }
    F0R(i,2) F0R(j,3) {
        swap(m[i][j],M[i][j]);
        M[i][j].clear();
    }
}

void solve() {
    cin >> n;
    string s; cin >> s;
    F0R(i,n-1) tmp[i] = s.substr(2*i,2);
    tmp[n-1] = s.substr(2*(n-1),3);
    F0R(i,n) cin >> num[i];
    
    F0R(i,2) F0R(j,3) m[i][j].clear();
    m[0][0][{0,0}] = 1;
    F0R(i,n) process(i);
    ll ans = 0;
    for (auto a: m[0][0]) ans += a.s;
    cout << ans << "\n";
}

int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin >> q;
    F0R(i,q) solve();
}

/* 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: 16 / 16
• Subtask 2: 0 / 17
• Subtask 3: 26 / 26
• Subtask 4: 0 / 23
• Subtask 5: 0 / 18