| # |
Submission time |
Handle |
Problem |
Language |
Result |
Execution time |
Memory |
| 62923 |
2018-07-30T20:52:39 Z |
Benq |
Spiral (BOI16_spiral) |
C++11 |
|
5 ms |
660 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;
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
*/
| # |
Verdict |
Execution time |
Memory |
Grader output |
| 1 |
Runtime error |
5 ms |
504 KB |
Execution killed with signal 11 (could be triggered by violating memory limits) |
| # |
Verdict |
Execution time |
Memory |
Grader output |
| 1 |
Runtime error |
3 ms |
644 KB |
Execution killed with signal 11 (could be triggered by violating memory limits) |
| # |
Verdict |
Execution time |
Memory |
Grader output |
| 1 |
Runtime error |
5 ms |
504 KB |
Execution killed with signal 11 (could be triggered by violating memory limits) |
| # |
Verdict |
Execution time |
Memory |
Grader output |
| 1 |
Correct |
5 ms |
660 KB |
Output is correct |
| # |
Verdict |
Execution time |
Memory |
Grader output |
| 1 |
Runtime error |
5 ms |
504 KB |
Execution killed with signal 11 (could be triggered by violating memory limits) |
