HDU 1402 A * B Problem Plus FFT

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【题目链接】点击打开链接
【题意】计算2个大数A*B的值，FFT入门题。记录一下FFT的模板，FFT的学习资料看这里：点击打开链接
【AC代码】
////Created by BLUEBUFF 2016/1/9//Copyright (c) 2016 BLUEBUFF.All Rights Reserved//#pragma comment(linker,"/STACK:102400000,102400000")#include <ext/pb_ds/assoc_container.hpp>#include <ext/pb_ds/tree_policy.hpp>#include <ext/pb_ds/hash_policy.hpp>#include <set>#include <map>#include <queue>#include <stack>#include <cmath>#include <cstdio>#include <time.h>#include <cstdlib>#include <cstring>#include <complex>#include <sstream> //isstringstream#include <iostream>#include <algorithm>using namespace std;//using namespace __gnu_pbds;typedef long long LL;typedef pair<int, LL> pp;#define REP1(i, a, b) for(int i = a; i < b; i++)#define REP2(i, a, b) for(int i = a; i <= b; i++)#define REP3(i, a, b) for(int i = a; i >= b; i--)#define CLR(a, b)     memset(a, b, sizeof(a))#define MP(x, y)      make_pair(x,y)template <class T1, class T2>inline void getmax(T1 &a, T2 b) { if (b>a)a = b; }template <class T1, class T2>inline void getmin(T1 &a, T2 b) { if (b<a)a = b; }const int maxn = 200010;const int maxm = 100010;const int maxs = 10;const int maxp = 1e3 + 10;const int INF  = 1e9;const int UNF  = -1e9;const int mod  = 1e9 + 7;const double PI = acos(-1);//headtypedef complex <double> Complex;void rader(Complex *y, int len) {    for(int i = 1, j = len / 2; i < len - 1; i++) {        if(i < j) swap(y[i], y[j]);        int k = len / 2;        while(j >= k) {j -= k; k /= 2;}        if(j < k) j += k;    }}void fft(Complex *y, int len, int op) {    rader(y, len);    for(int h = 2; h <= len; h <<= 1) {        double ang = op * 2 * PI / h;        Complex wn(cos(ang), sin(ang));        for(int j = 0; j < len; j += h) {            Complex w(1, 0);            for(int k = j; k < j + h / 2; k++) {                Complex u = y[k];                Complex t = w * y[k + h / 2];                y[k] = u + t;                y[k + h / 2] = u - t;                w = w * wn;            }        }    }    if(op == -1) for(int i = 0; i < len; i++) y[i] /= len;}Complex x1[maxn], x2[maxn];char str1[maxn], str2[maxn];int sum[maxn];int main(){    while(scanf("%s%s", str1, str2) != EOF)    {        int len1 = strlen(str1);        int len2 = strlen(str2);        int len = 1;        while(len < len1 * 2 || len < len2 * 2) len <<= 1;        for(int i = 0; i < len1; i++) x1[i] = Complex(str1[len1 - 1 - i] - '0', 0);        for(int i = len1; i < len; i++) x1[i] = Complex(0, 0);        for(int i = 0; i < len2; i++) x2[i] = Complex(str2[len2 - i - 1] - '0', 0);        for(int i = len2; i < len; i++) x2[i] = Complex(0, 0);        //DFT        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++) sum[i] = (int) (x1[i].real() + 0.5);        for(int i = 0; i < len; i++){            sum[i + 1] += sum[i] / 10;            sum[i] %= 10;        }        len = len1 + len2 - 1;        while(sum[len] <= 0 && len > 0) len--;        for(int i = len; i >= 0; i--) printf("%c", sum[i] + '0');        printf("\n");    }    return 0;}
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