CODEVS 3123 高精度练习之超大整数乘法

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题目描述 Description

给出两个正整数A和B，计算A*B的值。保证A和B的位数不超过100000位。

输入描述 Input Description

读入两个用空格隔开的正整数

输出描述 Output Description

输出A*B的值

样例输入 Sample Input

4 9

样例输出 Sample Output

数据范围及提示 Data Size & Hint

两个正整数的位数不超过100000位

分析

FFT板子题

代码

#include <bits/stdc++.h>using namespace std;const double _2pi = 3.1415926535 * 2;const int maxn = 10000000;int ans[maxn];struct complexNumber{    double real, image;};complexNumber add(complexNumber a, complexNumber b){    return (complexNumber){a.real + b.real, a.image + b.image};}complexNumber subtract(complexNumber a, complexNumber b){    return (complexNumber){a.real - b.real, a.image - b.image};}complexNumber multiply(complexNumber a, complexNumber b){    return (complexNumber){a.real * b.real - a.image * b.image, a.real * b.image + a.image * b.real};}vector<complexNumber> A, B;void init(){    ios::sync_with_stdio(false);    string n, m;    cin >> n >> m;    for(int i = n.length() - 1; i >= 0; i--)        A.push_back((complexNumber){double(n[i] - '0'), 0.0});    for(int i = m.length() - 1; i >= 0; i--)        B.push_back((complexNumber){double(m[i] - '0'), 0.0});}vector<complexNumber> FFT(vector<complexNumber> X, bool inverse){    int n = X.size();    for(int i = 0, j = 0; i < n; i++)    {        if(i < j)            swap(X[i], X[j]);        int k = n;        while(j & (k >>= 1))            j &= ~k;        j |= k;    }    double theta = inverse == false ? _2pi : -_2pi;    for(int s = 1; s <= log2((double)n); s++)    {        int m = pow(2, s);        complexNumber omega_n = (complexNumber){cos(theta / double(m)), sin(theta / double(m))};        for(int k = 0; k < n; k += m)        {            complexNumber omega = (complexNumber){1.0, 0.0};            for(int j = 0; j < m / 2; j++)            {                complexNumber u = X[k + j], t = multiply(omega, X[k + j + m / 2]);                X[k + j] = add(u, t);                X[k + j + m / 2] = subtract(u, t);                omega = multiply(omega, omega_n);            }        }    }    if(inverse == true)        for(int i = 0; i < n; i++)            X[i] = multiply(X[i], (complexNumber){1.0 / n, 0.0});    return X;}vector<complexNumber> Convolution(vector<complexNumber> A, vector<complexNumber> B){    vector<complexNumber> C;    int s1 = A.size(), s2 = B.size(), n = 2;    while(n < s1 + s2)        n <<= 1;    for(int i = s1; i < n; i++)        A.push_back((complexNumber){0.0, 0.0});    vector<complexNumber> alpha = FFT(A, false);    for(int i = s2; i < n; i++)        B.push_back((complexNumber){0.0, 0.0});    vector<complexNumber> beta = FFT(B, false);    for(int i = 0; i < n; i++)        C.push_back(multiply(alpha[i], beta[i]));    C = FFT(C, true);    return C;}void solve(){    vector<complexNumber> C = Convolution(A, B);    int len = A.size() + B.size() - 1;    for(int i = 0; i < len; i++)        ans[i] = (int)round(C[i].real);    for(int i = 0; i < len; i++)        ans[i + 1] += ans[i] / 10, ans[i] %= 10;    while(ans[len] == 0 && len > 0)        len--;    for(int i = len; i >= 0; i--)        cout << ans[i];}int main(){    init();    solve();    return 0;}

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