Java, lớp số nguyên lớn tùy chỉnh: 32.9 (120000000/365000)
Lớp chính khá đơn giản:
import java.util.*;
public class PPCG37270 {
public static void main(String[] args) {
long start = System.nanoTime();
int n = 12000000;
if (args.length == 1) n = Integer.parseInt(args[0]);
boolean[] sieve = new boolean[n + 1];
int[] remaining = new int[n + 1];
int[] count = new int[n + 1];
for (int p = 2; p <= n; p++) {
if (sieve[p]) continue;
long p2 = p * (long)p;
if (p2 > n) continue;
for (int i = (int)p2; i <= n; i += p) sieve[i] = true;
}
for (int i = 2; i <= n; i++) remaining[i] = i;
for (int p = 2; p <= n; p++) {
if (sieve[p]) continue;
for (int i = p; i <= n; i += p) {
while (remaining[i] % p == 0) {
remaining[i] /= p;
count[p]++;
if (i <= n/2) count[p] -= 2;
}
}
}
count[2] -= count[5];
count[5] = 0;
List<BigInt> partialProd = new ArrayList<BigInt>();
long accum = 1;
for (int i = 2; i <= n; i++) {
for (int j = count[i]; j > 0; j--) {
long tmp = accum * i;
if (tmp < 1000000000L) accum = tmp;
else {
partialProd.add(new BigInt((int)accum));
accum = i;
}
}
}
partialProd.add(new BigInt((int)accum));
System.out.println(prod(partialProd).digsum());
System.out.println((System.nanoTime() - start) / 1000000 + "ms");
}
private static BigInt prod(List<BigInt> vals) {
while (vals.size() > 1) {
int n = vals.size();
List<BigInt> next = new ArrayList<BigInt>();
for (int i = 0; i < n; i += 2) {
if (i == n - 1) next.add(vals.get(i));
else next.add(vals.get(i).mul(vals.get(i+1)));
}
vals = next;
}
return vals.get(0);
}
}
Nó dựa vào một lớp số nguyên lớn được tối ưu hóa cho phép nhân và toString()cả hai đều là những tắc nghẽn đáng kể trong quá trình thực hiện java.math.BigInteger.
/**
* A big integer class which is optimised for conversion to decimal.
* For use in simple applications where BigInteger.toString() is a bottleneck.
*/
public class BigInt {
// The base of the representation.
private static final int B = 1000000000;
// The number of decimal digits per digit of the representation.
private static final int LOG10_B = 9;
public static final BigInt ZERO = new BigInt(0);
public static final BigInt ONE = new BigInt(1);
// We use sign-magnitude representation.
private final boolean negative;
// Least significant digit is at val[off]; most significant is at val[off + len - 1]
// Unless len == 1 we guarantee that val[off + len - 1] is non-zero.
private final int[] val;
private final int off;
private final int len;
// Toom-style multiplication parameters from
// Zuras, D. (1994). More on squaring and multiplying large integers. IEEE Transactions on Computers, 43(8), 899-908.
private static final int[][][] Q = new int[][][]{
{},
{},
{{1, -1}},
{{4, 2, 1}, {1, 1, 1}, {1, 2, 4}},
{{8, 4, 2, 1}, {-8, 4, -2, 1}, {1, 1, 1, 1}, {1, -2, 4, -8}, {1, 2, 4, 8}}
};
private static final int[][][] R = new int[][][]{
{},
{},
{{1, -1, 1}},
{{-21, 2, -12, 1, -6}, {7, -1, 10, -1, 7}, {-6, 1, -12, 2, -21}},
{{-180, 6, 2, -80, 1, 3, -180}, {-510, 4, 4, 0, -1, -1, 120}, {1530, -27, -7, 680, -7, -27, 1530}, {120, -1, -1, 0, 4, 4, -510}, {-180, 3, 1, -80, 2, 6, -180}}
};
private static final int[][] S = new int[][]{
{},
{},
{1, 1, 1},
{1, 6, 2, 6, 1},
{1, 180, 120, 360, 120, 180, 1}
};
/**
* Constructs a big version of an integer value.
* @param x The value to represent.
*/
public BigInt(int x) {
this(Integer.toString(x));
}
/**
* Constructs a big version of a long value.
* @param x The value to represent.
*/
public BigInt(long x) {
this(Long.toString(x));
}
/**
* Parses a decimal representation of an integer.
* @param str The value to represent.
*/
public BigInt(String str) {
this(str.charAt(0) == '-', split(str));
}
/**
* Constructs a sign-magnitude representation taking the entire span of the array as the range of interest.
* @param neg Is the value negative?
* @param val The base-B digits, least significant first.
*/
private BigInt(boolean neg, int[] val) {
this(neg, val, 0, val.length);
}
/**
* Constructs a sign-magnitude representation taking a range of an array as the magnitude.
* @param neg Is the value negative?
* @param val The base-B digits, least significant at offset off, most significant at off + val - 1.
* @param off The offset within the array.
* @param len The number of base-B digits.
*/
private BigInt(boolean neg, int[] val, int off, int len) {
// Bounds checks
if (val == null) throw new IllegalArgumentException("val");
if (off < 0 || off >= val.length) throw new IllegalArgumentException("off");
if (len < 1 || off + len > val.length) throw new IllegalArgumentException("len");
this.negative = neg;
this.val = val;
this.off = off;
// Enforce the invariant that this.len is 1 or val[off + len - 1] is non-zero.
while (len > 1 && val[off + len - 1] == 0) len--;
this.len = len;
// Sanity check
for (int i = 0; i < len; i++) {
if (val[off + i] < 0) throw new IllegalArgumentException("val contains negative digits");
}
}
/**
* Splits a string into base-B digits.
* @param str The string to parse.
* @return An array which can be passed to the (boolean, int[]) constructor.
*/
private static int[] split(String str) {
if (str.charAt(0) == '-') str = str.substring(1);
int[] arr = new int[(str.length() + LOG10_B - 1) / LOG10_B];
int i, off;
// Each element of arr represents LOG10_B characters except (probably) the last one.
for (i = 0, off = str.length() - LOG10_B; off > 0; off -= LOG10_B) {
arr[i++] = Integer.parseInt(str.substring(off, off + LOG10_B));
}
arr[i] = Integer.parseInt(str.substring(0, off + LOG10_B));
return arr;
}
public boolean isZero() {
return len == 1 && val[off] == 0;
}
public BigInt negate() {
return new BigInt(!negative, val, off, len);
}
public BigInt add(BigInt that) {
// If the signs differ, then since we use sign-magnitude representation we want to do a subtraction.
boolean isSubtraction = negative ^ that.negative;
BigInt left, right;
if (len < that.len) {
left = that;
right = this;
}
else {
left = this;
right = that;
// For addition I just care about the lengths of the arrays.
// For subtraction I want the largest absolute value on the left.
if (isSubtraction && len == that.len) {
int cmp = compareAbsolute(that);
if (cmp == 0) return ZERO; // Cheap special case
if (cmp < 0) {
left = that;
right = this;
}
}
}
if (right.isZero()) return left;
BigInt result;
if (!isSubtraction) {
int[] sum = new int[left.len + 1];
// A copy here rather than using left.val in the main loops and copying remaining values
// at the end gives a small performance boost, probably due to cache locality.
System.arraycopy(left.val, left.off, sum, 0, left.len);
int carry = 0, k = 0;
for (; k < right.len; k++) {
int a = sum[k] + right.val[right.off + k] + carry;
sum[k] = a % B;
carry = a / B;
}
for (; carry > 0 && k < left.len; k++) {
int a = sum[k] + carry;
sum[k] = a % B;
carry = a / B;
}
sum[left.len] = carry;
result = new BigInt(negative, sum);
}
else {
int[] diff = new int[left.len];
System.arraycopy(left.val, left.off, diff, 0, left.len);
int carry = 0, k = 0;
for (; k < right.len; k++) {
int a = diff[k] - right.val[right.off + k] + carry;
// Why did anyone ever think that rounding positive and negative divisions differently made sense?
if (a < 0) {
diff[k] = a + B;
carry = -1;
}
else {
diff[k] = a % B;
carry = a / B;
}
}
for (; carry != 0 && k < left.len; k++) {
int a = diff[k] + carry;
if (a < 0) {
diff[k] = a + B;
carry = -1;
}
else {
diff[k] = a % B;
carry = a / B;
}
}
result = new BigInt(left.negative, diff, 0, k > left.len ? k : left.len);
}
return result;
}
private int compareAbsolute(BigInt that) {
if (len > that.len) return 1;
if (len < that.len) return -1;
for (int i = len - 1; i >= 0; i--) {
if (val[off + i] > that.val[that.off + i]) return 1;
if (val[off + i] < that.val[that.off + i]) return -1;
}
return 0;
}
public BigInt mul(BigInt that) {
if (isZero() || that.isZero()) return ZERO;
if (len == 1) return that.mulSmall(negative ? -val[off] : val[off]);
if (that.len == 1) return mulSmall(that.negative ? -that.val[that.off] : that.val[that.off]);
int shorter = len < that.len ? len : that.len;
BigInt result;
// Cutoffs have been hand-tuned.
if (shorter > 300) result = mulToom(3, that);
else if (shorter > 28) result = mulToom(2, that);
else result = mulNaive(that);
return result;
}
BigInt mulSmall(int m) {
if (m == 0) return ZERO;
if (m == 1) return this;
if (m == -1) return negate();
// We want to do the magnitude calculation with a positive multiplicand.
boolean neg = negative;
if (m < 0) {
neg = !neg;
m = -m;
}
int[] pr = new int[len + 1];
int carry = 0;
for (int i = 0; i < len; i++) {
long t = val[off + i] * (long)m + carry;
pr[i] = (int)(t % B);
carry = (int)(t / B);
}
pr[len] = carry;
return new BigInt(neg, pr);
}
// NB This truncates.
BigInt divSmall(int d) {
if (d == 0) throw new ArithmeticException();
if (d == 1) return this;
if (d == -1) return negate();
// We want to do the magnitude calculation with a positive divisor.
boolean neg = negative;
if (d < 0) {
neg = !neg;
d = -d;
}
int[] div = new int[len];
int rem = 0;
for (int i = len - 1; i >= 0; i--) {
long t = val[off + i] + rem * (long)B;
div[i] = (int)(t / d);
rem = (int)(t % d);
}
return new BigInt(neg, div);
}
BigInt mulNaive(BigInt that) {
int[] rv = new int[len + that.len];
// Naive multiplication
for (int i = 0; i < len; i++) {
for (int j = 0; j < that.len; j++) {
int k = i + j;
long c = val[off + i] * (long)that.val[that.off + j];
while (c > 0) {
c += rv[k];
rv[k] = (int)(c % B);
c /= B;
k++;
}
}
}
return new BigInt(this.negative ^ that.negative, rv);
}
private BigInt mulToom(int k, BigInt that) {
// We split each number into k parts of m base-B digits each.
// m = ceil(longer / k)
int m = ((len > that.len ? len : that.len) + k - 1) / k;
// Perform the splitting and evaluation steps of Toom-Cook.
BigInt[] f1 = this.toomFwd(k, m);
BigInt[] f2 = that.toomFwd(k, m);
// Pointwise multiplication.
for (int i = 0; i < f1.length; i++) f1[i] = f1[i].mul(f2[i]);
// Inverse (or interpolation) and recomposition.
return toomBk(k, m, f1, negative ^ that.negative, val[off], that.val[that.off]);
}
// Splits a number into k parts of m base-B digits each and does the polynomial evaluation.
private BigInt[] toomFwd(int k, int m) {
// Split.
BigInt[] a = new BigInt[k];
for (int i = 0; i < k; i++) {
int o = i * m;
if (o >= len) a[i] = ZERO;
else {
int l = m;
if (o + l > len) l = len - o;
// Ignore signs for now.
a[i] = new BigInt(false, val, off + o, l);
}
}
// Evaluate
return transform(Q[k], a);
}
private BigInt toomBk(int k, int m, BigInt[] f, boolean neg, int lsd1, int lsd2) {
// Inverse (or interpolation).
BigInt[] b = transform(R[k], f);
// Recomposition: add at suitable offsets, dividing by the normalisation factors
BigInt prod = ZERO;
int[] s = S[k];
for (int i = 0; i < b.length; i++) {
int[] shifted = new int[i * m + b[i].len];
System.arraycopy(b[i].val, b[i].off, shifted, i * m, b[i].len);
prod = prod.add(new BigInt(neg ^ b[i].negative, shifted).divSmall(s[i]));
}
// Handle the remainders.
// In the worst case the absolute value of the sum of the remainders is s.length, so pretty small.
// It should be easy enough to work out whether to go up or down.
int lsd = (int)((lsd1 * (long)lsd2) % B);
int err = lsd - prod.val[prod.off];
if (err > B / 2) err -= B / 2;
if (err < -B / 2) err += B / 2;
return prod.add(new BigInt(err));
}
/**
* Multiplies a matrix of small integers and a vector of big ones.
* The matrix has a implicit leading row [1 0 ... 0] and an implicit trailing row [0 ... 0 1].
* @param m The matrix.
* @param v The vector.
* @return m v
*/
private BigInt[] transform(int[][] m, BigInt[] v) {
BigInt[] b = new BigInt[m.length + 2];
b[0] = v[0];
for (int i = 0; i < m.length; i++) {
BigInt s = ZERO;
for (int j = 0; j < m[i].length; j++) s = s.add(v[j].mulSmall(m[i][j]));
b[i + 1] = s;
}
b[b.length - 1] = v[v.length - 1];
return b;
}
/**
* Sums the digits of this integer.
* @return The sum of the digits of this integer.
*/
public long digsum() {
long rv = 0;
for (int i = 0; i < len; i++) {
int x = val[off + i];
while (x > 0) {
rv += x % 10;
x /= 10;
}
}
return rv;
}
}
Nút thắt lớn là phép nhân ngây thơ (60%), tiếp theo là phép nhân khác (37%) và sàng (3%). Cuộc digsum()gọi không đáng kể.
Hiệu suất được đo bằng OpenJDK 7 (64 bit).