WeChall - Valentine's Gold

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Challenge

题目定义了 <USERNAME>'s number：一个偶数可以写成两个质数之和时，满足 p <= q 的质数对 (p, q) 数量。例如：

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F(4)  = 1  (2+2)
F(6)  = 1  (3+3)
F(8)  = 1  (3+5)
F(10) = 2  (3+7, 5+5)
F(12) = 1  (5+7)

需要计算 F(4), F(6), ..., F(31415926)，把这些十进制数字直接拼接成一个字符串，再计算整个字符串的 MD5。

Solution

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/* Valentine's Gold - Goldbach partition counting via NTT (Number Theoretic Transform)
 *
 * For even n from 4 to N=31415926, count Goldbach pairs (p<=q, p+q=n, both prime).
 * The convolution coefficient is at most the number of primes up to N, so one NTT
 * modulus is sufficient: the exact coefficient is far below MOD and cannot wrap.
 *
 * NTT prime: p = 469762049 = 7 * 2^26 + 1, primitive root 3.
 * It supports the required transform length 2^26 and is larger than every coefficient.
 */
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <openssl/evp.h>

#define N 31415926
/* FFT size must be > 2*N, use power of 2 */
#define FFT_SIZE (1 << 26)  /* 67108864 > 2*N */

/* NTT parameters */
#define MOD 469762049LL
#define G 3LL

typedef long long ll;

ll pw(ll a, ll b, ll m) {
    ll r = 1; a %= m;
    while (b > 0) {
        if (b & 1) r = r * a % m;
        a = a * a % m;
        b >>= 1;
    }
    return r;
}

void ntt(ll *a, int n, int invert) {
    for (int i = 1, j = 0; i < n; i++) {
        int bit = n >> 1;
        for (; j & bit; bit >>= 1) j ^= bit;
        j ^= bit;
        if (i < j) { ll t = a[i]; a[i] = a[j]; a[j] = t; }
    }
    for (int len = 2; len <= n; len <<= 1) {
        ll w = invert ? pw(G, MOD - 1 - (MOD - 1) / len, MOD) : pw(G, (MOD - 1) / len, MOD);
        for (int i = 0; i < n; i += len) {
            ll wn = 1;
            for (int j = 0; j < len / 2; j++) {
                ll u = a[i + j], v = a[i + j + len / 2] * wn % MOD;
                a[i + j] = (u + v) % MOD;
                a[i + j + len / 2] = (u - v + MOD) % MOD;
                wn = wn * w % MOD;
            }
        }
    }
    if (invert) {
        ll ninv = pw(n, MOD - 2, MOD);
        for (int i = 0; i < n; i++) a[i] = a[i] * ninv % MOD;
    }
}

int main() {
    fprintf(stderr, "Sieving up to %d...\n", N);
    char *is_prime = calloc(N + 1, 1);
    memset(is_prime + 2, 1, N - 1);
    int sqrtN = (int)sqrt((double)N) + 1;
    for (int i = 2; i <= sqrtN; i++) {
        if (is_prime[i]) {
            for (ll j = (ll)i * i; j <= N; j += i)
                is_prime[j] = 0;
        }
    }
    fprintf(stderr, "Sieve done.\n");

    /* Build NTT input */
    fprintf(stderr, "Allocating NTT array (size %d, ~512MB)...\n", FFT_SIZE);
    ll *a = malloc(sizeof(ll) * FFT_SIZE);
    if (!a) { fprintf(stderr, "malloc failed for a\n"); return 1; }
    for (int i = 0; i < FFT_SIZE; i++) a[i] = 0;
    for (int i = 2; i <= N; i++)
        if (is_prime[i]) a[i] = 1;

    /* Forward NTT */
    fprintf(stderr, "NTT forward...\n");
    ntt(a, FFT_SIZE, 0);

    /* Square in frequency domain */
    fprintf(stderr, "Squaring...\n");
    for (int i = 0; i < FFT_SIZE; i++)
        a[i] = a[i] * a[i] % MOD;

    /* Inverse NTT */
    fprintf(stderr, "NTT backward...\n");
    ntt(a, FFT_SIZE, 1);

    /* Extract Goldbach pair counts for even n */
    /* a[n] = convolution = number of ordered pairs (p,q) with p+q=n, both prime */
    /* For unordered pairs p<=q: G(n) = (a[n] + is_prime[n/2]) / 2 */
    fprintf(stderr, "Building result string...\n");

    size_t buf_size = 150 * 1024 * 1024;
    char *result = malloc(buf_size);
    size_t pos = 0;

    for (int n = 4; n <= N; n += 2) {
        ll ordered = a[n];
        ll half = is_prime[n / 2] ? 1 : 0;
        ll g = (ordered + half) / 2;
        pos += sprintf(result + pos, "%lld", g);
    }
    result[pos] = '\0';
    fprintf(stderr, "Result length: %zu\n", pos);
    fprintf(stderr, "First 50: %.50s\n", result);
    fprintf(stderr, "Last 50: %.50s\n", result + pos - 50);

    /* Verify */
    fprintf(stderr, "G(4)=%lld G(6)=%lld G(8)=%lld G(10)=%lld G(12)=%lld\n",
            (a[4]+is_prime[2])/2, (a[6]+is_prime[3])/2, (a[8]+is_prime[4])/2,
            (a[10]+is_prime[5])/2, (a[12]+is_prime[6])/2);

    /* MD5 */
    unsigned char md5_result[16];
    unsigned int md5_len;
    EVP_MD_CTX *mdctx = EVP_MD_CTX_new();
    EVP_DigestInit_ex(mdctx, EVP_md5(), NULL);
    EVP_DigestUpdate(mdctx, result, pos);
    EVP_DigestFinal_ex(mdctx, md5_result, &md5_len);
    EVP_MD_CTX_free(mdctx);

    printf("MD5: ");
    for (int i = 0; i < 16; i++) printf("%02x", md5_result[i]);
    printf("\n");

    free(result); free(a); free(is_prime);
    return 0;
}
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$ gcc -O2 -o goldbach_ntt goldbach_ntt.c -lm -lssl -lcrypto -march=native
$ ./goldbach_ntt