整数乘法，你能做多少位？

整数相乘

问题：在几秒内求出两个正整数的乘积？

这个简单的问题，你能做到哪一步？

如果两个5位整数相乘，这样就可以了

#include <cstdio>

int main() {
    int a, b;
    scanf("%d%d", &a, &b);
    printf("%d\n", a*b);
    return 0;
}

但如果要计算10位的整数相乘，那就得改成这样了

#include <cstdio>

int main() {
    long long a, b;
    scanf("%lld%lld", &a, &b);
    printf("%lld\n", a*b);
    return 0;
}

如果整数上百位千位了，那就要这样了

#include <iostream>
#include <string>
#include <vector>
#include <stdlib.h>
using namespace std;

void reverse_data(string &data);
string multiply(string numa, string numb);

void reverse_data(string &data) {
    char temp = '0';
    int start=0;
    int end=data.size()-1;
    while (start < end) {
        temp = data[start];
        data[start++] = data[end];
        data[end--] = temp;
    }
}

string multiply(string numa, string numb) {
    reverse_data(numa);
    reverse_data(numb);
    int i, j, c = 0;
    string temp(numa.size()+numb.size(),'0');
    for (i=0;i<numa.size();i++) {
        for (j=0;j<numb.size();j++) {
            c+=temp[i+j]-'0'+(numa[i]-'0')*(numb[j]-'0');
            temp[i+j]=(c%10)+'0';
            c=c/10;
        }
        while(c) {
            temp[i+j++]+=c%10;
            c=c/10;
        }
    }
    for (i=temp.size()-1;i>=0;i--) {
        if (temp[i] != '0') {
            break;
        }
    }
    temp = i >= 0 ? temp.substr(0, i+1): "0";
    reverse_data(temp);
    return temp;
}

int main() {
    string a,b;
    cin >>a >>b;
    string result = multiply(a, b);
    cout << result <<endl;
    return 0;
}

而要相乘的整数达到上百万位的时候，就得这样了。

#include <stdio.h>
#include <math.h>
#include <string.h>
const int BASE = 100000;
const int N_DIGIT = 5;
const int N = 32768*16;
const double PI = acos(-1.0);

struct plex {
    double real, imag;
};

plex omega[N >> 1];
plex a[N];
plex b[N];

char s[1000003];
int d[N], len;

void bitReverse(plex a[]) {
    int i, j = 1, k;
    plex t;
    for (i = 1; i < len; ++ i)
    {
        if (i < j)
        {
            t.real = a[j - 1].real;
            t.imag = a[j - 1].imag;
            a[j - 1].real = a[i - 1].real;
            a[j - 1].imag = a[i - 1].imag;
            a[i - 1].real = t.real;
            a[i - 1].imag = t.imag;
        }
        k = len >> 1;
        while (k < j)
        {
            j -= k;
            k >>= 1;
        }
        j += k;
    }
}

void calOmega() {
    double unit = 2 * PI / len;
    int n = len >> 1;
    for (int i = 0; i < n; ++ i)
    {
        double t = unit * i;
        omega[i].real = cos( t );
        omega[i].imag = sin( t );
    }
}

void fft(plex a[], bool inverse = false){
    bitReverse( a );

    int s = len >> 1;
    int m, k, j;
    int up, t, step;
    int i1, i2;
    plex tmp;

    if ( inverse ) {
        for (j = 0; j < s; ++ j)
            omega[j].imag = - omega[j].imag;
    }
    s = 1;
    for (m = 2; m <= len; m <<= 1) {
        up = m >> 1, t = len >> s; // 2^(log2(n) - s) != n - 2^s !!!!!!!
        for (k = 0; k < len; k += m) {
            step = 0;
            for (j = 0; j < up; ++ j) {
                i1 = k + j;
                i2 = i1 + up;
                tmp.real = omega[step].real * a[i2].real - omega[step].imag * a[i2].imag;
                tmp.imag = omega[step].real * a[i2].imag + omega[step].imag * a[i2].real;
                a[i2].real = a[i1].real - tmp.real;
                a[i2].imag = a[i1].imag - tmp.imag;
                a[i1].real += tmp.real;
                a[i1].imag += tmp.imag;
                step += t;
            }
        }
        ++ s;
    }
    if ( inverse ) {
        double t = 1.0 / len;
        for (j = 0; j < len; ++ j)
            a[j].real *= t;
    }
}

int convert(int d[], char s[])
{
    int sLen = strlen( s );
    int dLen = ((sLen - 1) / N_DIGIT) + 1, i = 0, n;
    char *pRight = s + sLen - 1, *pLeft = pRight - (N_DIGIT - 1);
    memset(d, 0, sizeof(int) * dLen);


    while (i < dLen && pRight >= s)
    {
        if (pLeft < s) pLeft = s;
        n = 0;
        while (pLeft <= pRight)
        {
            n = n * 10 + (*pLeft & 15);
            ++ pLeft;
        }
        d[i ++] = n;
        pRight -= N_DIGIT;
        pLeft = pRight - (N_DIGIT - 1);
    }
    return dLen;
}

bool init() {
    int i, j;
    if (scanf("%s", s) != 1)
        return false;
    int aLen = convert(d, s); //length of a
    for (i = 0; i < aLen; ++ i) {
        a[i].real = d[i];
        a[i].imag = 0;
    }

    scanf("%s", s);
    int bLen = convert(d, s);
    for (j = 0; j < bLen; ++ j)
    {
        b[j].real = d[j];
        b[j].imag = 0;
    }

    len = 1; //length of product who uses int
    while (len < aLen + bLen) len <<= 1;

    memset(a + i, 0, sizeof(plex) * (len - i));
    memset(b + j, 0, sizeof(plex) * (len - j));

    calOmega();
    return true;
}

void mul()
{
    for (int i = 0; i < len; ++ i)
    {
        double real = a[i].real * b[i].real - a[i].imag * b[i].imag;
        double imag = a[i].real * b[i].imag + a[i].imag * b[i].real;
        a[i].real = real;
        a[i].imag = imag;
    }
}

void print() {
    double carry = 0, t;
    static char format[10];
    int i;

    for (i = 0; i < len; ++ i) {
        t = carry + a[i].real;
        carry = floor((t + 0.5) / BASE);
        d[i] = int(t - carry * BASE + 0.5);
    }
    for (i = len - 1; i > 0 && d[i] == 0; -- i);
    sprintf(format, "%%.%dd", N_DIGIT);
    printf("%d", d[i]);
    for (-- i; i >= 0; -- i)
        printf(format, d[i]);
    printf("\n");
}

int main() {
    int inn; scanf( "%d", &inn );
    while ( inn-- ) {
        init();
        fft( a );
        fft( b );
        mul();
        fft(a, true);
        print();
    }
    return 0;
}

作者原创，转载请注明出处