This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include <bits/stdc++.h>
using namespace std;
#define int __int128
using ll = long long;
int modulo = 1e9 + 7;
void testcase()
{
ll X1,Y1,X2,Y2;
cin >> X1 >> Y1 >> X2 >> Y2;
int x1 = X1,x2 = X2,y1 = Y1,y2 = Y2;
if (x1 == 1 and y1 == 1)
{
int x = x2,y = y2,d = min(x2,y2);
int s = -d + 2 * d * d * (d + 1) * (d + 1) - 2 * d * (d + 1) * (2 * d + 1) + 3 * d * (d + 1) + 4 * d * (d + 1) * (2 * d + 1) / 6 - d * (d + 1);
s %= modulo;
if (x == y)
cout << (ll)s << '\n';
else if (x > y)
{
int s2 = (x - y) * y * (y + 1) / 2 + y * x * (x + 1) / 2 - y * y * (y + 1) / 2 + y * (x - y) + 4 * y * x * (x + 1) * (2 * x + 1) / 6 - 4 * y * y * (y + 1) * (2 * y + 1) / 6 - 4 * y * x * (x + 1) / 2 + 4 * y * y * (y + 1) / 2;
s += s2;
s %= modulo;
cout << (ll)s << '\n';
}
else
{
int s2 = x * (y - x) + 4 * x * y * (y + 1) * (2 * y + 1) / 6 - 4 * x * x * (x + 1) * (2 * x + 1) / 6 + 2 * x * x * (x + 1) - 2 * x * y * (y + 1) + x * y * (y + 1) - x * x * (x + 1) - (y - x) * x * x + x * y * (y + 1) / 2 - x * x * (x + 1) / 2 + (y - x) * x * (x - 1) / 2;
s += s2;
s %= modulo;
cout << (ll)s << '\n';
}
}
}
signed main()
{
ll n,tc;
cin >> n >> tc;
while (tc--)
testcase();
return 0;
}
/**
cea mai de mate de a 5a problema
**/
/**
idee: hai sa luam subtaskul ala de x1 = y1 = 1
Daca ar fi un patrat d * d, formula (doamne ajuta sa fi dat bine) este:
S = -d + 2 * d * d * (d + 1) * (d + 1) - 2 * d * (d + 1) * (2 * d + 1) + 3 * d * (d + 1) + 4 * d * (d + 1) * (2 * d + 1) / 6 - d * (d + 1)
Daca X > Y, mai am un S2 = (x - y) * y * (y + 1) / 2 + y * x * (x + 1) / 2 - y * y * (y + 1) / 2 + y * (x - y) + 4 * y * x * (x + 1) * (2 * x + 1) / 6 - 4 * y * (y + 1) * (2 * y + 1) / 6 - 4 * y * x * (x + 1) / 2 + 4 * y * (y + 1) / 2
Daca X < Y, mai am un S2 = x * (y - x) + 4 * x * y * (y + 1) * (2 * y + 1) / 6 - 4 * x * x * (x + 1) * (2 * x + 1) / 6 + 4 * x * x * (x + 1) / 2 - 4 * x * y * (y + 1) / 2 + x * y * (y + 1) - x * x * (x + 1) - (y * x) * x * x + x * y * (y + 1) / 2 - x * x * (x + 1) / 2 + (y - x) * x * (x - 1) / 2
Care-i sansa sa nu fi gresit niciun calcul?
Upd: 0
**/
# | Verdict | Execution time | Memory | Grader output |
---|
Fetching results... |
# | Verdict | Execution time | Memory | Grader output |
---|
Fetching results... |
# | Verdict | Execution time | Memory | Grader output |
---|
Fetching results... |
# | Verdict | Execution time | Memory | Grader output |
---|
Fetching results... |
# | Verdict | Execution time | Memory | Grader output |
---|
Fetching results... |