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 LL long long int
/*
(ex0, ye0) -> Given coordinate with flatten grass
(ex1, ye1) -> The lower-left coordinate of the block where (ex0, ye0) belongs to
(ex2, ye2) -> The lower-left coordinate of the whole structure
*/
int m;
LL n, ex0, ye0, ex1, ye1, ex2, ye2;
int smallestJumpToUnflattenedInDirection[4];
int testingMatrix[100][100];
// up, right, down, left
int dx[] = {0, 1, 0, -1};
int dy[] = {1, 0, -1, 0};
bool isInsideGrid(LL x, LL y) {
return x >= 1 && x <= n && y >= 1 && y <= n;
}
bool examine(int x, int y) {
cout << "examine " << x << " " << y << endl;
cout << flush;
string response;
cin >> response;
return response == "true";
}
void solution(int x, int y) {
cout << "solution " << x << " " << y << endl;
cout << flush;
}
void solve() {
for (int direction = 0; direction < 4; direction++) {
int threshold;
for (int p2 = 1; ; p2 *= 2) {
LL x = ex0 + (LL) p2 * dx[direction], y = ye0 + (LL) p2 * dy[direction];
if (!isInsideGrid(x, y) || examine(x, y) == false) {
threshold = p2;
break;
}
}
int lo = 1, hi = threshold, mid;
while (hi >= lo) {
mid = (lo + hi)/2;
LL x = ex0 + (LL) mid * dx[direction], y = ye0 + (LL) mid * dy[direction];
if (isInsideGrid(x, y) && examine(x, y)) {
lo = mid + 1;
} else {
hi = mid - 1;
}
}
smallestJumpToUnflattenedInDirection[direction] = lo;
}
m = (ye0 + smallestJumpToUnflattenedInDirection[0]) - (ye0 - smallestJumpToUnflattenedInDirection[2]) - 1;
ex1 = (ex0 - smallestJumpToUnflattenedInDirection[3]) + 1;
ye1 = (ye0 - smallestJumpToUnflattenedInDirection[2]) + 1;
int cntBlocksToTheLeft = 0;
for (int i = 1; ; i++) {
if (isInsideGrid(ex1 - i * 2 * m, ye1) && examine(ex1 - i * 2 * m, ye1)) {
cntBlocksToTheLeft += 1;
} else {
break;
}
}
int cntBlocksBelow = 0;
for (int i = 1; ; i++) {
if (isInsideGrid(ex1, ye1 - i * 2 * m) && examine(ex1, ye1 - i * 2 * m)) {
cntBlocksBelow += 1;
} else {
break;
}
}
ex2 = ex1 - cntBlocksToTheLeft * 2 * m;
ye2 = ye1 - cntBlocksBelow * 2 * m;
solution(ex2 + 2 * m + (m + 1)/2 - 1, ye2 + 2 * m + (m + 1)/2 - 1);
}
int main() {
cin >> n >> ex0 >> ye0;
solve();
}
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