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SPOJ VFMUL - Very Fast Multiplication (FFT)

题目链接:VFMUL - Very Fast Multiplication

Description

Multiply the given numbers.

Input

n [the number of multiplications <= 101]

l1 l2 [numbers to multiply (at most 300000 decimal digits each)]

Text grouped in [ ] does not appear in the input file.

Output

The results of multiplications.

Example

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Input:
5
4 2
123 43
324 342
0 12
9999 12345

Output:
8
5289
110808
0
123437655

Warning: large Input/Output data, be careful with certain languages

Solution

题意

求两数的乘积

思路

FFT

FFT 求高精度乘法的模板题。

Code

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#include <bits/stdc++.h>
using namespace std;

const double PI = acos(-1);
typedef complex<double> Complex;
const int maxn = 3e6 + 10;

Complex a[maxn], b[maxn];
int m, n;
int bit = 2, rev[maxn];
int ans[maxn];

void get_rev(){
memset(rev, 0, sizeof(rev));
while(bit <= n + m) bit <<= 1;
for(int i = 0; i < bit; ++i) {
rev[i] = (rev[i >> 1] >> 1) | (bit >> 1) * (i & 1);
}
}

void FFT(Complex *a, int op) {
for(int i = 0; i < bit; ++i) {
if(i < rev[i]) swap(a[i], a[rev[i]]);
}
for(int mid = 1; mid < bit; mid <<= 1) {
Complex wn = Complex(cos(PI / mid), op * sin(PI / mid));
for(int j = 0; j < bit; j += mid<<1) {
Complex w(1, 0);
for(int k = 0; k < mid; ++k, w = w * wn) {
Complex x = a[j + k], y = w * a[j + k + mid];
a[j + k] = x + y, a[j + k + mid] = x - y;
}
}
}
}

int main() {
int T;
scanf("%d", &T);
while(T--) {
memset(a, 0, sizeof(a));
memset(b, 0, sizeof(b));
string s1, s2;
cin >> s1 >> s2;
n = s1.size(), m = s2.size();
for(int i = 0; i < n; ++i) {
a[i] = s1[n - i - 1] - '0';
}
for(int i = 0; i < m; ++i) {
b[i] = s2[m - i - 1] - '0';
}
get_rev();
FFT(a, 1);
FFT(b, 1);
for(int i = 0; i <= bit; ++i) {
a[i] *= b[i];
}
FFT(a, -1);
for(int i = 0; i < n + m; ++i) {
ans[i] = (int)(a[i].real() / bit + 0.5);
}
for(int i = 1; i < n + m; ++i) {
ans[i] = ans[i] + ans[i - 1] / 10;
ans[i - 1] = ans[i - 1] % 10;
}
int s = n + m - 1;
for(; s >= 0; --s) {
if(ans[s]) break;
}
if(s < 0) printf("0\n");
else {
for(int i = s; i >= 0; --i) {
printf("%d", ans[i]);
}
printf("\n");
}
}
return 0;
}

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