hdu 1402 A * B Problem Plus fft

题目链接

A * B Problem Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16382    Accepted Submission(s): 3325

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模板题。

 

#include <iostream>
#include <vector>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <queue>
#include <complex>
#include <stack>
#include <bitset>
using namespace std;
#define pb(x) push_back(x)
#define ll long long
#define mk(x, y) make_pair(x, y)
#define lson l, m, rt<<1
#define mem(a) memset(a, 0, sizeof(a))
#define rson m+1, r, rt<<1|1
#define mem1(a) memset(a, -1, sizeof(a))
#define mem2(a) memset(a, 0x3f, sizeof(a))
#define rep(i, n, a) for(int i = a; i<n; i++)
#define fi first
#define se second
typedef pair<int, int> pll;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int mod = 1e9+7;
const int inf = 1061109567;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
typedef complex <double> cmx;
void change(cmx y[],int len)
{
    int i,j,k;
    for(i = 1, j = len/2;i < len-1; i++)
    {
        if(i < j)
            swap(y[i], y[j]);
        k = len / 2;
        while( j >= k)
        {
            j -= k;
            k /= 2;
        }
        if(j < k) j += k;
    }
}
void fft(cmx y[],int len,int on)
{
    change(y,len);
    for(int h = 2; h <= len; h <<= 1)
    {
        cmx wn(cos(-on*2*PI/h),sin(-on*2*PI/h));
        for(int j = 0;j < len;j+=h)
        {
            cmx w(1,0);
            for(int k = j;k < j+h/2;k++)
            {
                cmx u = y[k];
                cmx t = w*y[k+h/2];
                y[k] = u+t;
                y[k+h/2] = u-t;
                w = w*wn;
            }
        }
    }
    if(on == -1)
        for(int i = 0;i < len;i++)
            y[i] /= len;
}
const int maxn = 200010;
int ans[maxn];
cmx x1[maxn], x2[maxn];
char s1[50005], s2[50005];
int main()
{
    while(~scanf("%s%s", s1, s2)) {
        int len1 = strlen(s1);
        int len2 = strlen(s2);
        mem(ans);
        int len = 1;
        while(len<len1*2 || len<len2*2)
            len<<=1;
        for(int i = 0; i<len1; i++) {
            x1[i] = cmx(s1[len1-i-1]-'0', 0);
        }
        for(int i = len1; i<len; i++)
            x1[i] = cmx(0, 0);
        for(int i = 0; i<len2; i++) {
            x2[i] = cmx(s2[len2-i-1]-'0', 0);
        }
        for(int i = len2; i<len; i++)
            x2[i] = cmx(0, 0);
        fft(x1, len, 1);
        fft(x2, len, 1);
        for(int i = 0; i<len; i++)
            x1[i] = x1[i]*x2[i];
        fft(x1, len, -1);
        for(int i = 0; i<len; i++)
            ans[i] = (int)(x1[i].real() + 0.5);
        for(int i = 0; i<len; i++) {
            ans[i+1] += ans[i]/10;
            ans[i]%=10;
        }
        len = len1+len2-1;
        while(len>0&&ans[len]==0)
            len--;
        for(int i = len; i>=0; i--)
            printf("%d", ans[i]);
        cout<<endl;
    }
    return 0;
}