C++ 函数查找Pi的第n位

我一直想找到一种算法来做到这一点。我不在乎它有多慢，只要它能返回π的第n位：

例：

最好不要使用无限级数

如果任何人都有这样的函数或类，C或C++中我会真正感兴趣的。p> Thank

演示了如何在O（N）时间和O（log·N）空间中计算π的第N位，并且不必计算其前面的所有数字

哦，是十六进制的

如果您不想这样做，您可以很容易地从shell执行此操作：

% perl -Mbignum=bpi -wle 'print bpi(20)'
3.1415926535897932385

% perl -Mbignum=bpi -wle 'print bpi(50)'
3.1415926535897932384626433832795028841971693993751

% perl -Mbignum=bpi -wle 'print bpi(200)'
3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679821480865132823066470938446095505822317253594081284811174502841027019385211055596446229489549303820

% perl -Mbignum=bpi -wle 'print bpi(1000)'
3.141592653589793238462643383279502884197169399375105820974944592307816406286208998628034825342117067982148086513282306647093844609550582231725359408128481117450284102701938521105559644622948954930381964428810975665933446128475648233786783165271201909145648566923460348610454326648213393607260249141273724587006606315588174881520920962829254091715364367892590360011330530548820466521384146951941511609433057270365759591953092186117381932611793105118548074462379962749567351885752724891227938183011949129833673362440656643086021394946395224737190702179860943702770539217176293176752384674818467669405132000568127145263560827785771342757789609173637178721468440901224953430146549585371050792279689258923542019956112129021960864034418159813629774771309960518707211349999998372978049951059731732816096318595024459455346908302642522308253344685035261931188171010003137838752886587533208381420617177669147303598253490428755468731159562863882353787593751957781857780532171226806613001927876611195909216420199

---

下面是由Fabrice Bellard编写的Simon Plouffe的解决方案：

   /*
     * Computation of the n'th decimal digit of \pi with very little memory.
     * Written by Fabrice Bellard on January 8, 1997.
     * 
     * We use a slightly modified version of the method described by Simon
     * Plouffe in "On the Computation of the n'th decimal digit of various
     * transcendental numbers" (November 1996). We have modified the algorithm
     * to get a running time of O(n^2) instead of O(n^3log(n)^3).
     * 
     * This program uses mostly integer arithmetic. It may be slow on some
     * hardwares where integer multiplications and divisons must be done
     * by software. We have supposed that 'int' has a size of 32 bits. If
     * your compiler supports 'long long' integers of 64 bits, you may use
     * the integer version of 'mul_mod' (see HAS_LONG_LONG).  
     */

    #include <stdlib.h>
    #include <stdio.h>
    #include <math.h>

/* uncomment the following line to use 'long long' integers */
/* #define HAS_LONG_LONG */

#ifdef HAS_LONG_LONG
#define mul_mod(a,b,m) (( (long long) (a) * (long long) (b) ) % (m))
#else
#define mul_mod(a,b,m) fmod( (double) a * (double) b, m)
#endif

/* return the inverse of x mod y */
int inv_mod(int x, int y)
{
    int q, u, v, a, c, t;

    u = x;
    v = y;
    c = 1;
    a = 0;
    do {
    q = v / u;

    t = c;
    c = a - q * c;
    a = t;

    t = u;
    u = v - q * u;
    v = t;
    } while (u != 0);
    a = a % y;
    if (a < 0)
    a = y + a;
    return a;
}

/* return (a^b) mod m */
int pow_mod(int a, int b, int m)
{
    int r, aa;

    r = 1;
    aa = a;
    while (1) {
    if (b & 1)
        r = mul_mod(r, aa, m);
    b = b >> 1;
    if (b == 0)
        break;
    aa = mul_mod(aa, aa, m);
    }
    return r;
}

/* return true if n is prime */
int is_prime(int n)
{
    int r, i;
    if ((n % 2) == 0)
    return 0;

    r = (int) (sqrt(n));
    for (i = 3; i <= r; i += 2)
    if ((n % i) == 0)
        return 0;
    return 1;
}

/* return the prime number immediatly after n */
int next_prime(int n)
{
    do {
    n++;
    } while (!is_prime(n));
    return n;
}

int main(int argc, char *argv[])
{
    int av, a, vmax, N, n, num, den, k, kq, kq2, t, v, s, i;
    double sum;

    if (argc < 2 || (n = atoi(argv[1])) <= 0) {
    printf("This program computes the n'th decimal digit of \\pi\n"
           "usage: pi n , where n is the digit you want\n");
    exit(1);
    }

    N = (int) ((n + 20) * log(10) / log(2));

    sum = 0;

    for (a = 3; a <= (2 * N); a = next_prime(a)) {

    vmax = (int) (log(2 * N) / log(a));
    av = 1;
    for (i = 0; i < vmax; i++)
        av = av * a;

    s = 0;
    num = 1;
    den = 1;
    v = 0;
    kq = 1;
    kq2 = 1;

    for (k = 1; k <= N; k++) {

        t = k;
        if (kq >= a) {
        do {
            t = t / a;
            v--;
        } while ((t % a) == 0);
        kq = 0;
        }
        kq++;
        num = mul_mod(num, t, av);

        t = (2 * k - 1);
        if (kq2 >= a) {
        if (kq2 == a) {
            do {
            t = t / a;
            v++;
            } while ((t % a) == 0);
        }
        kq2 -= a;
        }
        den = mul_mod(den, t, av);
        kq2 += 2;

        if (v > 0) {
        t = inv_mod(den, av);
        t = mul_mod(t, num, av);
        t = mul_mod(t, k, av);
        for (i = v; i < vmax; i++)
            t = mul_mod(t, a, av);
        s += t;
        if (s >= av)
            s -= av;
        }

    }

    t = pow_mod(10, n - 1, av);
    s = mul_mod(s, t, av);
    sum = fmod(sum + (double) s / (double) av, 1.0);
    }
    printf("Decimal digits of pi at position %d: %09d\n", n,
       (int) (sum * 1e9));
    return 0;
}

---

包括此解决方案在内的多个解决方案实际上打印了从第n位开始的多个数字

现在有一个很好的解决方案（比以前的答案快得多）

代码只需要将“typedef int64 ModInt；”行修改为“typedef int64_t ModInt；”

---

这是“2015年初”macbook pro上的一个问题。

这似乎是（一年前你自己的问题）的完全复制品。作为一个新问题，你再次提出这个问题有什么原因吗？@Mu Mind这是一个公式，但现在我要求实现。@jmasterx在我的回答中特别看到bullet#3…你不希望它返回

int

，还是

无符号int

，米洛：你在如何实现它的问题上有没有具体的问题？这不是WriteMap程序。com@Milo：如果您不理解公式，无法进行编程，那么您没有从中学习任何东西，并且可能有实际需要：也许可以通过从文件中读取pi值来满足它（您可以从“net-e.g.”下载该值，在你感兴趣的职位上建立索引？为您节省大量计算时间…；-）以16为基数计算第n位与以10为基数计算第n位有很大不同。从一个到另一个在时间和空间上都有不平凡的大O！这个答案与OP的问题无关；该链接描述了如何找到第n个基数-16位数字，而问题意味着基数为10。每个找到此页面的人在这里都有一个原因：他们有实际需要。虽然此链接可能会回答问题，但最好在这里包含答案的基本部分，并提供链接供参考。如果链接页面发生更改，仅链接的答案可能无效。这不会提供问题的答案。若要评论或要求作者澄清，请在他们的帖子下方留下评论。@abcaefchen：我正在研究，但我对论文的解释在我的程序中还不起作用。@king.code:Milo要求一个给出π的第n个小数点的函数。这篇论文包含了一个算法，它正好做到了这一点，这将是编写c程序的第一步！回答得好。我想衡量一下实现离O（n²）有多近。

   /*
     * Computation of the n'th decimal digit of \pi with very little memory.
     * Written by Fabrice Bellard on January 8, 1997.
     * 
     * We use a slightly modified version of the method described by Simon
     * Plouffe in "On the Computation of the n'th decimal digit of various
     * transcendental numbers" (November 1996). We have modified the algorithm
     * to get a running time of O(n^2) instead of O(n^3log(n)^3).
     * 
     * This program uses mostly integer arithmetic. It may be slow on some
     * hardwares where integer multiplications and divisons must be done
     * by software. We have supposed that 'int' has a size of 32 bits. If
     * your compiler supports 'long long' integers of 64 bits, you may use
     * the integer version of 'mul_mod' (see HAS_LONG_LONG).  
     */

    #include <stdlib.h>
    #include <stdio.h>
    #include <math.h>

/* uncomment the following line to use 'long long' integers */
/* #define HAS_LONG_LONG */

#ifdef HAS_LONG_LONG
#define mul_mod(a,b,m) (( (long long) (a) * (long long) (b) ) % (m))
#else
#define mul_mod(a,b,m) fmod( (double) a * (double) b, m)
#endif

/* return the inverse of x mod y */
int inv_mod(int x, int y)
{
    int q, u, v, a, c, t;

    u = x;
    v = y;
    c = 1;
    a = 0;
    do {
    q = v / u;

    t = c;
    c = a - q * c;
    a = t;

    t = u;
    u = v - q * u;
    v = t;
    } while (u != 0);
    a = a % y;
    if (a < 0)
    a = y + a;
    return a;
}

/* return (a^b) mod m */
int pow_mod(int a, int b, int m)
{
    int r, aa;

    r = 1;
    aa = a;
    while (1) {
    if (b & 1)
        r = mul_mod(r, aa, m);
    b = b >> 1;
    if (b == 0)
        break;
    aa = mul_mod(aa, aa, m);
    }
    return r;
}

/* return true if n is prime */
int is_prime(int n)
{
    int r, i;
    if ((n % 2) == 0)
    return 0;

    r = (int) (sqrt(n));
    for (i = 3; i <= r; i += 2)
    if ((n % i) == 0)
        return 0;
    return 1;
}

/* return the prime number immediatly after n */
int next_prime(int n)
{
    do {
    n++;
    } while (!is_prime(n));
    return n;
}

int main(int argc, char *argv[])
{
    int av, a, vmax, N, n, num, den, k, kq, kq2, t, v, s, i;
    double sum;

    if (argc < 2 || (n = atoi(argv[1])) <= 0) {
    printf("This program computes the n'th decimal digit of \\pi\n"
           "usage: pi n , where n is the digit you want\n");
    exit(1);
    }

    N = (int) ((n + 20) * log(10) / log(2));

    sum = 0;

    for (a = 3; a <= (2 * N); a = next_prime(a)) {

    vmax = (int) (log(2 * N) / log(a));
    av = 1;
    for (i = 0; i < vmax; i++)
        av = av * a;

    s = 0;
    num = 1;
    den = 1;
    v = 0;
    kq = 1;
    kq2 = 1;

    for (k = 1; k <= N; k++) {

        t = k;
        if (kq >= a) {
        do {
            t = t / a;
            v--;
        } while ((t % a) == 0);
        kq = 0;
        }
        kq++;
        num = mul_mod(num, t, av);

        t = (2 * k - 1);
        if (kq2 >= a) {
        if (kq2 == a) {
            do {
            t = t / a;
            v++;
            } while ((t % a) == 0);
        }
        kq2 -= a;
        }
        den = mul_mod(den, t, av);
        kq2 += 2;

        if (v > 0) {
        t = inv_mod(den, av);
        t = mul_mod(t, num, av);
        t = mul_mod(t, k, av);
        for (i = v; i < vmax; i++)
            t = mul_mod(t, a, av);
        s += t;
        if (s >= av)
            s -= av;
        }

    }

    t = pow_mod(10, n - 1, av);
    s = mul_mod(s, t, av);
    sum = fmod(sum + (double) s / (double) av, 1.0);
    }
    printf("Decimal digits of pi at position %d: %09d\n", n,
       (int) (sum * 1e9));
    return 0;
}

C:\tcc>tcc -I libtcc libtcc/libtcc.def -run examples/pi.c 1
Decimal digits of pi at position 1: 141592653

C:\tcc>tcc -I libtcc libtcc/libtcc.def -run examples/pi.c 1000
Decimal digits of pi at position 1000: 938095257

gcc pidec.cpp && time echo 20000 | ./a.out
Pidec, direct computation of decimal digits of pi at a given position n.
(http://numbers.computation.free.fr/Constants/constants.html for more details)
Enter n : Parameters : M=122, N=7094, M*N+M=872562
Series time : 0.17
Digits of pi after n-th decimal digit : 203856539
Total time: 0.68

real    0m0.691s
user    0m0.678s
sys 0m0.006s

Compilation finished at Wed Feb  3 13:34:48