제출 #62923

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
62923		Benq	Spiral (BOI16_spiral)	C++11	31 / 100	5 ms	660 KiB

spiral

#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;


ll po (ll b, ll p) { return !p?1:po(b*b%MOD,p/2)*(p&1?b:1)%MOD; }
ll inv (ll b) { return po(b,MOD-2); }

int ad(int a, int b) { return (a+b)%MOD; }
int sub(int a, int b) { return (a-b+MOD)%MOD; }
int mul(int a, int b) { return (ll)a*b%MOD; }

int AD(int& a, int b) { return a = ad(a,b); }
int SUB(int& a, int b) { return a = sub(a,b); }
int MUL(int& a, int b) { return a = mul(a,b); }

int eval2(int x) {
    return mul(mul(mul(x,ad(x,1)),ad(mul(2,x),1)),inv(6));
}

int getSum(int l, int r, int deg) {
    if (deg == 0) return ad(sub(r,l),1);
    if (deg == 1) return mul(mul(ad(l,r),ad(sub(r,l),1)),inv(2));
    if (deg == 2) return sub(eval2(r),eval2(l-1));
    exit(5);
}

vi operator+(const vi& l, const vi& r) {
    vi res(max(sz(l),sz(r)));
    F0R(i,sz(l)) res[i] = l[i];
    F0R(i,sz(r)) AD(res[i],r[i]);
    return res;
}
vi operator-(const vi& l, const vi& r) {
    vi res(max(sz(l),sz(r)));
    F0R(i,sz(l)) res[i] = l[i];
    F0R(i,sz(r)) SUB(res[i],r[i]);
    return res;
}
vi operator*(const vi& l, const vi& r) {
    vi x(sz(l)+sz(r)-1);
    F0R(i,sz(l)) F0R(j,sz(r)) AD(x[i+j],mul(l[i],r[j]));
    return x;
}
vi operator*(const vi& l, const int& r) {
    vi L = l;
    for (int& i: L) MUL(i,r);
    return L;
}

vi operator+=(vi& l, const vi& r) { return l = l+r; }
vi operator-=(vi& l, const vi& r) { return l = l-r; }
template<class T> vi operator*=(vi& l, const T& r) { return l = l*r; }

std::ostream& operator<<(std::ostream &strm, const vi& a) {
    cout << "{";
    F0R(i,sz(a)) {
        if (i) cout << ", ";
        cout << a[i];
    }
    cout << "}\n";
    return strm;
}
    
vi interpolate(vpi v) {
    vi ret;
    F0R(i,sz(v)) {
        vi prod = {1};
        int todiv = 1;
        F0R(j,sz(v)) if (i != j) {
            MUL(todiv,sub(v[i].f,v[j].f));
            vi tmp = {sub(0,v[j].f),1};
            prod *= tmp;
        }
        ret += prod*mul(inv(todiv),v[i].s);
    }
    return ret;
}

int eval(vi v, int y) {
    vpi part; part.pb({0,0});
    F0R(i,sz(v)) part.pb({i+1,ad(part.back().s,v[i])});
    
    vi z = interpolate(part);
    int cur = 1, ret = 0;
    F0R(i,sz(z)) {
        AD(ret,mul(z[i],cur));
        MUL(cur,y);
    }
    
    return ret;
}

int get(int a, int b) {
    if (a == 0 && b == 0) return 1;
    int r = max(abs(a),abs(b));
    if (a == r && b > -r) return (2*r-1)*(2*r-1)+r+b;
    if (b == r) return (2*r-1)*(2*r-1)+3*r-a;
    
    if (a == -r) return (2*r-1)*(2*r-1)+5*r-b;
    return (2*r-1)*(2*r-1)+7*r+a;
}

int n,q;

int brute(int x1, int x2, int y1, int y2) {
    int ans = 0;
    FOR(i,x1,x2+1) FOR(j,y1,y2+1) ans += get(i,j);
    return ans;
}

int get1(int x, int y) {
    // y > 0, y <= x
    // (2*a-1)*(2*a-1)+a+b;
    y = min(y,x);
    vi v;
    FOR(Y,1,min(y,4)+1) {
        int t = mul(4,getSum(Y,x,2));
        t = sub(t,mul(3,getSum(Y,x,1)));
        t = ad(t,mul(ad(1,Y),getSum(Y,x,0)));
        v.pb(t);
    }
    return eval(v,y);
}

int get2(int x, int y) { 
    // y>x, x >= 0
    // (2*r-1)*(2*r-1)+3*r-a;
    x = min(x,y);
    vi v;
    F0R(X,min(x+1,4)) {
        int t = mul(4,getSum(X+1,y,2));
        t = sub(t,getSum(X+1,y,1));
        // cout << t << "\n";
        t = ad(t,mul(getSum(X+1,y,0),sub(1,X)));
        /*cout << t << "\n";
        cout << "AH " << X+1 << " " << y << " " << t << "\n";*/
        v.pb(t);
    }
    return eval(v,x+1);
}

int solve(int x1, int x2, int y1, int y2) {
    assert(x1 == 1 && y1 == 1);
    // cout << get2(x2,y2) << "\n";
    int ans = sub(ad(get1(x2,y2),get2(x2,y2)),get2(0,y2));
   //  cout << ans << " " << brute(x1,x2,y1,y2) << "\n";
    // assert(ans == brute(x1,x2,y1,y2));
    return ans;
}

int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin >> n >> q;
    F0R(i,q) {
        int x1, y1, x2, y2; cin >> x1 >> y1 >> x2 >> y2;
        cout << solve(x1,x2,y1,y2) << "\n";
    }
    /*cout << get(1,0) << "\n";
    FORd(j,-5,6) {
        FOR(i,-5,6) cout << get(i,j) << "\t";
        cout << "\n";
    }*/
}

/* 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: 0 / 12
• Subtask 2: 0 / 15
• Subtask 3: 0 / 17
• Subtask 4: 31 / 31
• Subtask 5: 0 / 25