Submission #126572

#	Time	Username	Problem	Language	Result	Execution time	Memory
126572		choikiwon	Solitaire (JOI16_solitaire)	C++17	80 / 100	239 ms	262148 KiB

solitaire

#include<bits/stdc++.h>
using namespace std;

typedef pair<int, int> pii;

const int mod = 1e9 + 7;
const int MN = 6010;

int exp(int x, int n) {
    int ret = 1;
    while(n) {
        if(n & 1) ret = 1LL * ret * x % mod;
        x = 1LL * x * x % mod;
        n >>= 1;
    }
    return ret;
}
int inv(int x) {
    return exp(x, mod - 2);
}
int fact[MN], invf[MN];
int comb(int n, int k) {
    if(n < k) return 0;
    return 1LL * fact[n] * invf[k] % mod * invf[n - k] % mod;
}
int H(int n, int k) {
    if(k == 0) return 1;
    return comb(n + k - 1, k);
}

int N, L, R;
char G[3][MN];
int psum[MN];

int calc(int l, int r) {
    if(l > r) return 0;
    return psum[r] - (l? psum[l - 1] : 0);
}

int dp1(int t, int r, int pre);

vector<vector<int> > cc3[3][3];
int dp3(int a, int b, int r, int pre) {
    int &ret = cc3[a][b][r][pre];
    if(ret != -1) return ret;

    int tr = L + r;
    int rem = calc(L, tr - 1);

    if(rem < pre) return ret = 0;

    ret = 0;
    if(pre < rem) {
        ret += dp3(a, b, r, pre + 1);
        ret %= mod;
    }
    ret += 1LL * dp1(1, r - 1, pre) * fact[a] % mod * fact[b] % mod * H(pre + 1, a) % mod * H(rem - pre + 1, b) % mod;
    ret %= mod;
    return ret;
}

vector<vector<int> > cc2[3][3];
int dp2(int a, int b, int r, int pre) {
    if(pre < 0) return 0;
    int &ret = cc2[a][b][r][pre];
    if(ret != -1) return ret;

    int tr = L + r;
    int rem = calc(L, tr - 1);

    ret = 0;
    if(pre) {
        ret += dp2(a, b, r, pre - 1);
        ret %= mod;
    }
    ret += 1LL * dp1(0, r - 1, pre) * fact[a] % mod * fact[b] % mod * H(pre + 1, a) % mod * H(rem - pre + 1, b) % mod;
    ret %= mod;

    return ret;
}

vector<vector<int> > cc1[2];
int dp1(int t, int r, int pre) {
    if(r == -1) {
        if(L) return pre == 0;
        else return pre == 0 && t == 0;
    }
    int &ret = cc1[t][r][pre];
    if(ret != -1) return ret;

    int tr = r + L;
    int rem = calc(L, tr - 1);
    int cnt = (G[0][tr] == 'x') + (G[2][tr] == 'x');

    ret = 0;
    if(t) {
        ret += dp2(cnt, 0, r, min(rem, pre - 1 - cnt)) % mod;
        ret %= mod;
        return ret;
    }
    else {
        ret += dp2(cnt, 0, r, rem);
        ret %= mod;
        for(int i = 0; i < cnt; i++) {
            ret += 1LL * comb(cnt, i) * dp3(i, cnt - i, r, max(0, pre - i)) % mod;
            ret %= mod;
        }
        return ret;
    }
}

int proc(int l, int r) {
    L = l;
    R = r;

    int tot = 0;
    for(int i = 0; i < 3; i++) {
        for(int j = L; j <= R; j++) {
            tot += G[i][j] == 'x';
        }
    }

    cc1[0] = cc1[1] = vector<vector<int> >(R - L + 1, vector<int>(tot + 1, -1));

    for(int i = 0; i < 3; i++) {
        for(int j = 0; j < 3; j++) {
            cc2[i][j] = cc3[i][j] = vector<vector<int> >(R - L + 1, vector<int>(tot + 1, -1));
        }
    }

    int ret;
    if(r == N - 1) ret = dp1(0, R - L, tot);
    else ret = dp1(0, R - L, 0);

    //cout << l << ' ' << r << ' ' << ret << endl;

    return ret;
}

int main() {
    fact[0] = 1;
    for(int i = 1; i < MN; i++) {
        fact[i] = 1LL * fact[i - 1] * i % mod;
    }
    for(int i = 0; i < MN; i++) {
        invf[i] = inv(fact[i]);
    }

    scanf("%d", &N);

    int sum = 0;
    for(int i = 0; i < 3; i++) {
        scanf("\n");
        for(int j = 0; j < N; j++) {
            scanf("%c", &G[i][j]);
            if(G[i][j] == 'x') sum++;
        }
    }
    for(int i = 0; i < N; i++) {
        for(int j = 0; j < 3; j++) {
            psum[i] += G[j][i] == 'x';
        }
        if(i) psum[i] += psum[i - 1];
    }

    if(G[0][0] == 'x' || G[0][N - 1] == 'x' || G[2][0] == 'x' || G[2][N - 1] == 'x') {
        printf("0");
        return 0;
    }

    for(int i = 0; i < N - 1; i++) {
        if(G[0][i] == 'x' && G[0][i + 1] == 'x') {
            printf("0");
            return 0;
        }
        if(G[2][i] == 'x' && G[2][i + 1] == 'x') {
            printf("0");
            return 0;
        }
    }

    vector<int> ind;
    vector<pii> prob;
    int s = 0;
    for(int i = 0; i < N; i++) {
        if(G[1][i] == 'o') {
            if(s < i) {
                int cnt = 0;
                for(int j = s; j < i; j++) {
                    for(int k = 0; k < 3; k++) cnt += G[k][j] == 'x';
                }
                ind.push_back(cnt);
                prob.push_back(pii(s, i - 1));
            }
            s = i + 1;
        }
    }
    if(s < N) {
        int cnt = 0;
        for(int j = s; j < N; j++) {
            for(int k = 0; k < 3; k++) cnt += G[k][j] == 'x';
        }
        ind.push_back(cnt);
        prob.push_back(pii(s, N - 1));
    }

    int ans = fact[sum];
    for(int i = 0; i < ind.size(); i++) {
        ans = 1LL * ans * invf[ ind[i] ] % mod;
    }
    for(int i = 0; i < prob.size(); i++) {
        ans = 1LL * ans * proc(prob[i].first, prob[i].second) % mod;
    }
    printf("%d", ans);
}

Compilation message (stderr)

solitaire.cpp: In function 'int main()':
solitaire.cpp:208:22: warning: comparison between signed and unsigned integer expressions [-Wsign-compare]
     for(int i = 0; i < ind.size(); i++) {
                    ~~^~~~~~~~~~~~
solitaire.cpp:211:22: warning: comparison between signed and unsigned integer expressions [-Wsign-compare]
     for(int i = 0; i < prob.size(); i++) {
                    ~~^~~~~~~~~~~~~
solitaire.cpp:149:10: warning: ignoring return value of 'int scanf(const char*, ...)', declared with attribute warn_unused_result [-Wunused-result]
     scanf("%d", &N);
     ~~~~~^~~~~~~~~~
solitaire.cpp:153:14: warning: ignoring return value of 'int scanf(const char*, ...)', declared with attribute warn_unused_result [-Wunused-result]
         scanf("\n");
         ~~~~~^~~~~~
solitaire.cpp:155:18: warning: ignoring return value of 'int scanf(const char*, ...)', declared with attribute warn_unused_result [-Wunused-result]
             scanf("%c", &G[i][j]);
             ~~~~~^~~~~~~~~~~~~~~~

• Subtask 1: 10 / 10
• Subtask 2: 12 / 12
• Subtask 3: 20 / 20
• Subtask 4: 38 / 38
• Subtask 5: 0 / 20