大数乘法 poj 2389 ||大数乘法 hdu1402 FFT模板

poj 2389：

Bull Math

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 13694 Accepted: 7065

Description

Bulls are so much better at math than the cows. They can multiply huge integers together and get perfectly precise answers ... or so they say. Farmer John wonders if their answers are correct. Help him check the bulls' answers. Read in two positive integers (no more than 40 digits each) and compute their product. Output it as a normal number (with no extra leading zeros). 

FJ asks that you do this yourself; don't use a special library function for the multiplication.

Input

* Lines 1..2: Each line contains a single decimal number.

Output

* Line 1: The exact product of the two input lines

Sample Input

111111111111111111111111

Sample Output

12345679011110987654321

思路 ：对应位子上存放对应的积的和 代码比较短 不信你看看

code：

#include <iostream>#include <stdio.h>#include<string.h>#include<malloc.h>using namespace std;int s[100];void multiply(const char *a,const char *b){    int i,j,ca,cb;    //int *s;    ca=strlen(a);    cb=strlen(b);    //s=(int *)malloc(sizeof(int)*(ca+cb));    for (i=0; i<ca+cb; i++)        s[i]=0;    for (i=0; i<ca; i++)        for (j=0; j<cb; j++)            s[i+j+1]+=(a[i]-'0')*(b[j]-'0');    for (i=ca+cb-1; i>=0; i--) //进位        if (s[i]>=10)        {            s[i-1]+=s[i]/10;            s[i]%=10;        }    i=0;    for(i=0; i<ca+cb; i++)        if(s[i]!=0)            break;// 跳过头部0元素    if(i==ca+cb)    {        printf("0\n");        return ;    }    for(; i<ca+cb; i++)        cout<<s[i];    cout<<endl;}int main(){    char a[51],b[51];    while(scanf("%s%s",a,b)!=EOF)    {        multiply(a,b);    }    return 0;}

hdu 1402：点击打开链接

A * B Problem Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14823    Accepted Submission(s): 2830

Problem Description

Calculate A * B.

 

Input

Each line will contain two integers A and B. Process to end of file.

Note: the length of each integer will not exceed 50000.

 

Output

For each case, output A * B in one line.

 

Sample Input

1210002

 

Sample Output

22000

FFT模板：转载：题解链接

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <cmath>using namespace std;#define L(x) (1 << (x))const double PI = acos(-1.0);const int Maxn = 133015;double ax[Maxn], ay[Maxn], bx[Maxn], by[Maxn];char sa[Maxn/2],sb[Maxn/2];int sum[Maxn];int x1[Maxn],x2[Maxn];int revv(int x, int bits){    int ret = 0;    for (int i = 0; i < bits; i++)    {        ret <<= 1;        ret |= x & 1;        x >>= 1;    }    return ret;}void fft(double * a, double * b, int n, bool rev){    int bits = 0;    while (1 << bits < n) ++bits;    for (int i = 0; i < n; i++)    {        int j = revv(i, bits);        if (i < j)            swap(a[i], a[j]), swap(b[i], b[j]);    }    for (int len = 2; len <= n; len <<= 1)    {        int half = len >> 1;        double wmx = cos(2 * PI / len), wmy = sin(2 * PI / len);        if (rev) wmy = -wmy;        for (int i = 0; i < n; i += len)        {            double wx = 1, wy = 0;            for (int j = 0; j < half; j++)            {                double cx = a[i + j], cy = b[i + j];                double dx = a[i + j + half], dy = b[i + j + half];                double ex = dx * wx - dy * wy, ey = dx * wy + dy * wx;                a[i + j] = cx + ex, b[i + j] = cy + ey;                a[i + j + half] = cx - ex, b[i + j + half] = cy - ey;                double wnx = wx * wmx - wy * wmy, wny = wx * wmy + wy * wmx;                wx = wnx, wy = wny;            }        }    }    if (rev)    {        for (int i = 0; i < n; i++)            a[i] /= n, b[i] /= n;    }}int solve(int a[],int na,int b[],int nb,int ans[]){    int len = max(na, nb), ln;    for(ln=0; L(ln)<len; ++ln);    len=L(++ln);    for (int i = 0; i < len ; ++i)    {        if (i >= na) ax[i] = 0, ay[i] =0;        else ax[i] = a[i], ay[i] = 0;    }    fft(ax, ay, len, 0);    for (int i = 0; i < len; ++i)    {        if (i >= nb) bx[i] = 0, by[i] = 0;        else bx[i] = b[i], by[i] = 0;    }    fft(bx, by, len, 0);    for (int i = 0; i < len; ++i)    {        double cx = ax[i] * bx[i] - ay[i] * by[i];        double cy = ax[i] * by[i] + ay[i] * bx[i];        ax[i] = cx, ay[i] = cy;    }    fft(ax, ay, len, 1);    for (int i = 0; i < len; ++i)        ans[i] = (int)(ax[i] + 0.5);    return len;}int main(){    int l1,l2,l;    int i;    while(gets(sa))    {        gets(sb);        memset(sum, 0, sizeof(sum));        l1 = strlen(sa);        l2 = strlen(sb);        for(i = 0; i < l1; i++)            x1[i] = sa[l1 - i - 1]-'0';        for(i = 0; i < l2; i++)            x2[i] = sb[l2-i-1]-'0';        l = solve(x1, l1, x2, l2, sum);        for(i = 0; i<l || sum[i] >= 10; i++) // 进位        {            sum[i + 1] += sum[i] / 10;            sum[i] %= 10;        }        l = i;        while(sum[l] <= 0 && l>0)    l--; // 检索最高位        for(i = l; i >= 0; i--)    putchar(sum[i] + '0'); // 倒序输出        putchar('\n');    }    return 0;}