結果
| 問題 |
No.2440 Accuracy of Integer Division Approximate Functions
|
| ユーザー |
|
| 提出日時 | 2023-08-24 21:12:57 |
| 言語 | D (dmd 2.109.1) |
| 結果 |
AC
|
| 実行時間 | 685 ms / 2,000 ms |
| コード長 | 2,591 bytes |
| コンパイル時間 | 2,475 ms |
| コンパイル使用メモリ | 173,244 KB |
| 実行使用メモリ | 5,248 KB |
| 最終ジャッジ日時 | 2024-12-23 07:12:16 |
| 合計ジャッジ時間 | 11,877 ms |
|
ジャッジサーバーID (参考情報) |
judge4 / judge1 |
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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 1 |
| other | AC * 20 |
ソースコード
import std.conv, std.functional, std.range, std.stdio, std.string;
import std.algorithm, std.array, std.bigint, std.bitmanip, std.complex, std.container, std.math, std.mathspecial, std.numeric, std.regex, std.typecons;
import core.bitop;
class EOFException : Throwable { this() { super("EOF"); } }
string[] tokens;
string readToken() { for (; tokens.empty; ) { if (stdin.eof) { throw new EOFException; } tokens = readln.split; } auto token = tokens.front; tokens.popFront; return token; }
int readInt() { return readToken.to!int; }
long readLong() { return readToken.to!long; }
real readReal() { return readToken.to!real; }
bool chmin(T)(ref T t, in T f) { if (t > f) { t = f; return true; } else { return false; } }
bool chmax(T)(ref T t, in T f) { if (t < f) { t = f; return true; } else { return false; } }
int binarySearch(alias pred, T)(in T[] as) { int lo = -1, hi = cast(int)(as.length); for (; lo + 1 < hi; ) { const mid = (lo + hi) >> 1; (unaryFun!pred(as[mid]) ? hi : lo) = mid; } return hi; }
int lowerBound(T)(in T[] as, T val) { return as.binarySearch!(a => (a >= val)); }
int upperBound(T)(in T[] as, T val) { return as.binarySearch!(a => (a > val)); }
// floor(a / b)
Int divFloor(Int)(Int a, Int b) {
return a / b - (((a ^ b) < 0 && a % b != 0) ? 1 : 0);
}
// sum_{x=l}^r floor((a x + b) / m)
// m > 0
BigInt sumFloors(BigInt m, BigInt a, BigInt b, BigInt l, BigInt r)
in {
assert(m > 0, "sumFloors: m > 0 must hold");
}
do {
BigInt sum;
for (; l <= r; ) {
const q = divFloor(a, m);
const aa = a - q * m;
const ll = divFloor(aa * l + b, m) + 1;
const rr = divFloor(aa * r + b, m);
// TODO: overflow if ll or rr is big in recursion
sum += q * ((r + 1) * r / 2 - l * (l - 1) / 2) + r * rr - (l - 1) * (ll - 1) + (rr - ll + 1);
a = m; m = aa; l = -rr; r = -ll;
}
return sum;
}
void main() {
try {
for (; ; ) {
const numCases = readInt;
foreach (caseId; 0 .. numCases) {
BigInt N = readLong;
const BigInt D = readLong;
const BigInt M = readLong;
const S = readInt;
const T = BigInt(1) << S;
const denom = T - M * D;
if (denom != 0) {
chmin(N, D * T / abs(denom));
}
debug {
writeln([N, D, M, T]);
}
BigInt ans;
if (1 <= N) {
ans += sumFloors(D, BigInt(1), BigInt(0), BigInt(1), N);
ans -= sumFloors(T, M, BigInt(0), BigInt(1), N);
}
ans = abs(ans);
ans = N - ans;
writeln(ans);
}
}
} catch (EOFException e) {
}
}