P1919 FFT加速高精度乘法

P1919 FFT加速高精度乘法

传送门：https://www.luogu.org/problemnew/show/P1919

题意：

给出两个n位10进制整数x和y，你需要计算x*y。

题解：

对于十进制数我们可以将其转换成

\(a0*10^0+a1*10^1+a2*10^2...an*10^n\)

那么对于两个数，我们就可以求出两个的系数表示后得到a的点乘式和b的点乘式

最后得到的答案就是a和b的多项式的系数，这个问题O（n）扫一遍，

处理一下输出即可

代码：

#include <set>
#include <map>
#include <cmath>
#include <cstdio>
#include <string>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
typedef unsigned long long uLL;
#define ls rt<<1
#define rs rt<<1|1
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define bug printf("*********\n")
#define FIN freopen("input.txt","r",stdin);
#define FON freopen("output.txt","w+",stdout);
#define IO ios::sync_with_stdio(false),cin.tie(0)
#define debug1(x) cout<<"["<<#x<<" "<<(x)<<"]\n"
#define debug2(x,y) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<"]\n"
#define debug3(x,y,z) cout<<"["<<#x<<" "<<(x)<<" "<<#y<<" "<<(y)<<" "<<#z<<" "<<z<<"]\n"
const int maxn = 1e6 + 5;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const double Pi = acos(-1.0);
LL quick_pow(LL x, LL y) {
    LL ans = 1;
    while(y) {
        if(y & 1) {
            ans = ans * x % mod;
        } x = x * x % mod;
        y >>= 1;
    } return ans;
}
struct complex {
    double x, y;
    complex(double xx = 0, double yy = 0) {
        x = xx, y = yy;
    }
} a[maxn], b[maxn];
complex operator + (complex a, complex b) {
    return complex(a.x + b.x, a.y + b.y);
}
complex operator - (complex a, complex b) {
    return complex(a.x - b.x, a.y - b.y);
}
complex operator * (complex a, complex b) {
    return complex(a.x * b.x - a.y * b.y, a.x * b.y + a.y * b.x);
}

int n, m;
int l, r[maxn];
int limit = 1;
void fft(complex *A, int type) {
    for(int i = 0; i < limit; i++) {
        if(i < r[i]) swap(A[i], A[r[i]]);
    }
    for(int mid = 1; mid < limit; mid <<= 1) {
        complex Wn(cos(Pi / mid), type * sin(Pi / mid));
        for(int R = mid << 1, j = 0; j < limit; j += R) {
            complex w(1, 0);
            for(int k = 0; k < mid; k++, w = w * Wn) {
                complex x = A[j + k], y = w * A[j + mid + k];
                A[j + k] = x + y;
                A[j + k + mid] = x - y;
            }
        }
    }
}
int ans[maxn];
char numA[maxn], numB[maxn];
int main() {
#ifndef ONLINE_JUDGE
    FIN
#endif
    int n;
    while(scanf("%d", &n) != EOF) {
        scanf("%s %s", numA, numB);
        // debug3(n,numA,numB);
        int lena = 0;
        int lenb = 0;
        for(int i = n - 1; i >= 0; i--) {
            a[lena++].x = numA[i] - '0';
        }
        for(int i = n - 1; i >= 0; i--) {
            b[lenb++].x = numB[i] - '0';
        }
        while(limit < n + n) limit <<= 1, l++;
        for(int i = 0; i <= limit; i++) {
            r[i] = (r[i >> 1] >> 1) | ((i & 1) << (l - 1));
        }
        fft(a, 1);
        fft(b, 1);
        for(int i = 0; i <= limit; i++) {
            a[i] = a[i] * b[i];
        }
        fft(a, -1);
        int tot = 0;
        for(int i = 0; i <= limit; i++) {
            ans[i] += (int)(a[i].x / limit + 0.5);
            if(ans[i] >= 10) {
                ans[i + 1] += ans[i] / 10;
                ans[i] %= 10;
                limit += (i == limit);
            }
        }
        while(!ans[limit] && limit >= 1) limit--;
        limit++;
        while(--limit >= 0) cout << ans[limit];
    }
    return 0;
}