数论-FTT 和 NTT

FTT

NKOJ3071 高精度乘（输入 a, b ，输出两个数的积）

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

using namespace std;

typedef complex<double> CP;
typedef long long LL;

const int _N = 300005;
const double PI = asin(1)*2;

LL rev[_N], Ans[_N];
char str1[_N], str2[_N];
CP X[_N], Y[_N];

void GetRev(LL bit)
{
    for (LL i = 0; i < (1<<bit); ++i)
        rev[i] = (rev[i>>1]>>1) | ((i&1)<<(bit-1));
    return;
}

void FFT(CP *A, LL n, LL ty)
{
    LL i, k, len;
    for (i = 0; i < n; ++i)
        if (i < rev[i]) swap(A[i], A[rev[i]]);
        
    for (len = 1; len < n; len <<= 1) {
        CP wn = exp(CP(0, ty*PI/len));
        for (i = 0; i < n; i += len<<1) {
            CP wi(1, 0);
            for (k = i; k < i+len; ++k) {
                CP t0 = A[k], t1 = wi*A[k+len];
                A[k] = t0+t1, A[k+len] = t0-t1;
                wi *= wn;
            }
        }
    }
    if (ty == -1)
        for (i = 0; i < n; ++i) A[i] /= n;
    return;
}

bool Input(LL &len, char *str)
{
    char tt; bool flag = false; int i;
    while (((tt = getchar()) < '0' || tt > '9') && tt != '-');
    if (tt == '-') flag = true, i = -1;
    else str[0] = tt, i = 0;
    while ((tt = getchar()) >= '0' && tt <= '9') str[++i] = tt;
    len = i+1;
    return flag;
}

int main()
{
    LL flag1, flag2, l1, l2, i;
    flag1 = Input(l1, str1), flag2 = Input(l2, str2);
//    printf("

%s %s

 %lld %lld
", str1, str2, l1, l2);
    LL a = 1, x = 0;
    while (a < l1+l2-1) a <<= 1, ++x;
//    printf("

a = %lld, x = %lld

", a, x);
    for (i = 0; i < l1; ++i) X[i] = (double)(str1[l1-1-i]-'0');
    for (i = 0; i < l2; ++i) Y[i] = (double)(str2[l2-1-i]-'0');
    GetRev(x), FFT(X, a, 1), FFT(Y, a, 1);
    for (i = 0; i < a; ++i) X[i] *= Y[i];
    FFT(X, a, -1);
    for (i = 0; i < l1+l2; ++i)
        Ans[i] += (long long)(X[i].real() + 0.5), Ans[i+1] += Ans[i]/10, Ans[i] %= 10;
    for (i = l1+l2; i >= 0 && !Ans[i]; --i);
    if (i == -1) { printf("0
"); return 0; }
    if (flag1 ^ flag2) putchar('-');
    while (i >= 0) putchar(Ans[i] + '0'), --i;
    putchar('
');
    return 0;
}

NTT

1.数论阶和原根

相关文章：数论之原根 https://blog.csdn.net/fuyukai/article/details/50894609

ord_n a 表示当 a, n 互素时，在模 n 意义下，a 的数论阶，即满足 a^x Ξ 1 (mod n) 的最小正整数 x .

结论：(ord_n a) | x

推论：(ord_n a) | phi(x)

当 n > 0 且 ord_n a = phi(n) 时，a 为 n 的一个原根.

素数一定有原根，整数不一定. 当一个数 n 有原根时，它的原根数量是 phi(phi(n)) .

在模 n 意义下，n 的原根为 g，gⁱ 互不相同 , 0 <= i <= phi(n) - 1

这一点性质类似 FFT 中的单位根 w ，也就是 NTT 的基础。

求素数 n 的原根

从小到大枚举每一个 x | phi(n) , 判断是否满足 a^x Ξ 1 (mod n)，找到一个合理的 x 即为 n 的原根.

2.NTT操作

#include <stdio.h>
#include <algorithm>

using namespace std;

typedef long long LL;

const int _N = 300005;
const long long P = 998244353LL;
const long long G = 3LL;

LL rev[_N], Ans[_N];
char str1[_N], str2[_N];
LL X[_N], Y[_N];

void GetRev(LL bit)
{
    for (LL i = 0; i < (1<<bit); ++i)
        rev[i] = (rev[i>>1]>>1) | ((i&1)<<(bit-1));
    return;
}

LL Mont(LL t1, LL t2)
{
    t1 %= P;
    LL ans = 1;
    while (t2) {
        if (t2 & 1) ans = ans*t1%P;
        t2 >>= 1, t1 = t1*t1%P;
    }
    return ans;
}

void NTT(LL *A, LL n, LL ty)
{
    LL i, k, len;
    for (i = 0; i < n; ++i)
        if (i < rev[i]) swap(A[i], A[rev[i]]);
        
    for (len = 1; len < n; len <<= 1) {
        LL wn = Mont(G, (P-1)+ty*(P-1)/(len<<1));
        for (i = 0; i < n; i += len<<1) {
            LL wi = 1;
            for (k = i; k < i+len; ++k) {
                LL t0 = A[k], t1 = wi*A[k+len]%P;
                A[k] = t0+t1, A[k+len] = t0-t1;
                if ((A[k] = t0+t1) >= P) A[k] -= P;
                if ((A[k+len] = t0-t1) < 0) A[k+len] += P;
                wi = wi*wn%P;
            }
        }
    }
    if (ty == -1) {
        LL tmp = Mont(n, P-2);
        for (i = 0; i < n; ++i) A[i] = A[i]*tmp%P;
    }
    return;
}

bool Input(LL &len, char *str)
{
    char tt; bool flag = false; int i;
    while (((tt = getchar()) < '0' || tt > '9') && tt != '-');
    if (tt == '-') flag = true, i = -1;
    else str[0] = tt, i = 0;
    while ((tt = getchar()) >= '0' && tt <= '9') str[++i] = tt;
    len = i+1;
    return flag;
}

int main()
{
    LL flag1, flag2, l1, l2, i;
    flag1 = Input(l1, str1), flag2 = Input(l2, str2);
//    printf("

%s %s

 %lld %lld
", str1, str2, l1, l2);
    LL a = 1, x = 0;
    while (a < l1+l2-1) a <<= 1, ++x;
//    printf("

a = %lld, x = %lld

", a, x);
    for (i = 0; i < l1; ++i) X[i] = str1[l1-1-i]-'0';
    for (i = 0; i < l2; ++i) Y[i] = str2[l2-1-i]-'0';
    GetRev(x), NTT(X, a, 1), NTT(Y, a, 1);
    for (i = 0; i < a; ++i) X[i] = X[i]*Y[i]%P;
    NTT(X, a, -1);
    for (i = 0; i < l1+l2; ++i)
        Ans[i] += X[i], Ans[i+1] += Ans[i]/10, Ans[i] %= 10;
    for (i = l1+l2; i >= 0 && !Ans[i]; --i);
    if (i == -1) { printf("0
"); return 0; }
    if (flag1 ^ flag2) putchar('-');
    while (i >= 0) putchar(Ans[i] + '0'), --i;
    putchar('
');
    return 0;
}