Submission #1126233

#	Time	Username	Problem	Language	Result	Execution time	Memory
1126233		InvMOD	Intergalactic ship (IZhO19_xorsum)	C++20	100 / 100	1626 ms	13848 KiB

xorsum

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

#define fi first
#define se second
#define gcd __gcd
#define sz(v) (int) v.size()
#define pb push_back
#define pi pair<int,int>
#define all(v) (v).begin(), (v).end()
#define compact(v) (v).erase(unique(all(v)), (v).end())
#define FOR(i, a, b) for(int i = (a); i <= (b); i++)
#define dbg(x) "[" #x " = " << (x) << "]"
///#define int long long

using ll = long long;
using ld = long double;
using ull = unsigned long long;

template<typename T> bool ckmx(T& a, const T& b){if(a < b) return a = b, true; return false;}
template<typename T> bool ckmn(T& a, const T& b){if(a > b) return a = b, true; return false;}

const int N = 1e3 + 5;
const int QN = 1e5 + 1;

const int MOD = 1e9 + 7; // modulo
int add(int x, int y){
    return (x + y < MOD ? x + y : x + y - MOD);
}

int sub(int x, int y){
    return add(x - y, MOD);
}

int mul(int x, int y){
    return (1ll * x * y) % MOD;
}

int bpow(int x, int y){
    int res = 1;
    while(y){
        if(y&1) res = mul(res, x);
        x = mul(x, x);
        y >>= 1;
    }
    return res;
}

const int lg = 8;

int n, q, a[N], pw2[QN], ql[QN], qr[QN], qx[QN], save_case[4][N][N];

// Cach de chon so luong le hoac chan thang trong x thang
// Lay san 1 thak nhung thak con lai tu sinh tu diet :D
int comb(int x, int type){
    // Luon co 1 cach cho chan
    if(!x) return (type ^ 1);
    assert(pw2[x - 1] > 0);
    return pw2[x - 1];
}

void update_data(int _case, int bt1, int bt2, int x1, int y1, int x2, int y2){
    save_case[_case][x1][y1]++;
    save_case[_case][x1][y2 + 1]--;
    save_case[_case][x2 + 1][y1]--;
    save_case[_case][x2 + 1][y2 + 1]++;
    return;
}

void preprocess(){
    for(int _case = 0; _case < 3; _case++){
        //cout << "CASE: " << _case <<"\n";
          //      cout << dbg(bt1) << dbg(bt2) <<"\n";
        for(int l = 1; l <= n; l++){
            for(int r = 1; r <= n; r++){
                // Prefix sum: a[x][y] = a[x][y] + a[x - 1][y] + a[x][y - 1] - a[x - 1][y - 1]
                //int dbg_val = save_case[_case][bt1][bt2][l][r];
    //            cout << "Pre: " << dbg(l) << dbg(r) << dbg(dbg_val) <<"\n";
                save_case[_case][l][r] +=
                    save_case[_case][l][r - 1] +
                    save_case[_case][l - 1][r] -
                    save_case[_case][l - 1][r - 1];
                //dbg_val = save_case[_case][l][r];
      //          cout << "Do: " << dbg(l) << dbg(r) << dbg(dbg_val) <<"\n";
               // assert(dbg_val >= 0);
            }
                //    cout <<"\n";
                //cout <<"\n";
        }
        //cout <<"\n";
    }
    return;
}

void solve()
{
    cin >> n;
    for(int i = 1; i <= n; i++){
        cin >> a[i];
    }

    cin >> q;
    for(int i = 1; i <= q; i++){
        cin >> ql[i] >> qr[i] >> qx[i];
    }

    pw2[0] = 1;
    for(int i = 1; i <= q; i++)
        pw2[i] = mul(pw2[i -1], 2);

    int answer = 0;

    // Xet 2 bit
    for(int bt1 = 0; bt1 < lg; bt1++){
        for(int bt2 = 0; bt2 < lg; bt2++){
            for(int _case = 0; _case < 3; _case++){
                for(int i = 0; i <= n; i++){
                    for(int j = 0; j <= n; j++){
                        save_case[_case][i][j] = 0;
                    }
                }
            }

            for(int i = 1; i <= q; i++){
                // Fix mot thak random va so sanh voi mot thak random khac va 2 bit bt1, bt2 ->
                // co 3 case anh huong khi query update: (0, 1) | (1, 0) | (1, 1)

                int l = ql[i], r = qr[i], x = qx[i];
                if((x >> bt1 & 1) && (x >> bt2 & 1)){

                    // (i, L) -> (i, R) (L <= i <= R) thi se la case 2: (1, 1)
                    update_data(2, bt1, bt2, l, l, r, r);

                    // (i, L) -> (i, R) (1 <= i < L) thi se la case 0: (0, 1)
                    update_data(0, bt1, bt2, 1, l, l-1, r);

                    // (L, i) -> (R, i) (R+1 <= i <= n) thi se la case 1: (1, 0)
                    update_data(1, bt1, bt2, l, r+1, r, n);
                } // Huhuhuhuhuhuhuhuhuhuhuuhuhuhuhuhuhuhuhuhuhuhuhuhu sai xam l
                else if((x >> bt1 & 1)){
                    // do ko update bt2 nen kieu j cx la (1, 0) (case 1)
                    // nhx thak (L,R) -> (L,n)
                    update_data(1, bt1, bt2, l, l, r, n);
                }
                else if((x >> bt2 & 1)){
                    // tuong tu bt1
                    // nhx thak (1,R) -> (L,R)
                    update_data(0, bt1, bt2, 1, l, r, r);
                }
            }
            preprocess();

            // Xet 2 thak random
            for(int i = 1; i <= n; i++){
                for(int j = i; j <= n; j++){
                    // So cach chon 2 thak i,j

                    int bta = (a[i] >> bt1 & 1);
                    int btb = (a[j] >> bt2 & 1);

                    int sum_case = 0;
                    for(int case0 = 0; case0 < 2; case0++){
                        for(int case1 = 0; case1 < 2; case1++){
                            for(int case2 = 0; case2 < 2; case2++){
                                int nbta = (bta + case1 + case2) & 1;
                                int nbtb = (btb + case0 + case2) & 1;
                                assert(nbta == 1 || nbta == 0);
                                assert(nbtb == 1 || nbtb == 0);
                                if(nbta > 0 && nbtb > 0){
                                    // So cach chon case0 * case1 * case2
                                    sum_case = add(sum_case,
                                    mul(comb(save_case[0][i][j], case0),
                                    mul(comb(save_case[1][i][j], case1),
                                        comb(save_case[2][i][j], case2)
                                    )));
                                }
                            }
                        }
                    }
                    // Nhung thak (0, 0) con lai chon sao cung dc
                    int left = q - save_case[0][i][j]
                                - save_case[1][i][j]
                                - save_case[2][i][j];
                        assert(left >= 0);
                    sum_case = mul(sum_case, pw2[left]);

                    // i = j -> 1 else 2
                    int tot = (i == j ? 1 : 2);
                    tot = mul(tot, (1 << bt1));
                    tot = mul(tot, (1 << bt2));

                    assert(tot > 0);

                    int choosePairs = mul(i, n - j + 1);
                    int contrib = mul(tot, mul(sum_case, choosePairs));

                    /*
                    if(contrib) cout << dbg(i) << dbg(j) << dbg(bt1) << dbg(bt2) << dbg(tot) << dbg(sum_case) << dbg(contrib) << "\n";

                    if(i == 1 && j == 1 && bt1 == 1 && bt2 == 0){
                        cout << "DUMB: " << contrib <<" " << left <<" " << sum_case <<" " << choosePairs <<" " << tot <<"\n";
                    }
                    */
                    answer = add(answer, contrib);
                }
            }
        }
    }

    cout << answer <<"\n";
    return;
}

signed main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    #define name "InvMOD"
    if(fopen(name".INP", "r")){
        freopen(name".INP","r",stdin);
        freopen(name".OUT","w",stdout);
    }

    int t = 1; //cin >> t;
    while(t--) solve();
    return 0;
}

/* Debug Test
2
2 1
1
1 1 1

Answer: 40

2
1 2
1
2 2 2

Answer: 16

*/

Compilation message (stderr)

xorsum.cpp: In function 'int main()':
xorsum.cpp:222:16: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)' declared with attribute 'warn_unused_result' [-Wunused-result]
  222 |         freopen(name".INP","r",stdin);
      |         ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~
xorsum.cpp:223:16: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)' declared with attribute 'warn_unused_result' [-Wunused-result]
  223 |         freopen(name".OUT","w",stdout);
      |         ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~

• Subtask 1: 4 / 4
• Subtask 2: 5 / 5
• Subtask 3: 6 / 6
• Subtask 4: 9 / 9
• Subtask 5: 8 / 8
• Subtask 6: 11 / 11
• Subtask 7: 19 / 19
• Subtask 8: 30 / 30
• Subtask 9: 8 / 8