poj 2389：

Bull Math

Time Limit: 1000MS	Memory Limit: 65536K
Total 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

11111111111111
1111111111

Sample Output

12345679011110987654321

思路 ：對應位子上存放對應的積的和 程式碼比較短 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--) //進位
        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;// 跳過頭部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;
}
 

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 1 2 1000 2
Sample Output 2 2000

FFT模板：轉載：

題解連結

#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++) // 進位
        {
            sum[i + 1] += sum[i] / 10;
            sum[i] %= 10;
        }

        l = i;
        while(sum[l] <= 0 && l>0)    l--; // 檢索最高位
        for(i = l; i >= 0; i--)    putchar(sum[i] + '0'); // 倒序輸出
        putchar('\n');
    }
    return 0;
}