hdu 1402 FFT

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就是求两个长度为50000的数相乘。
第一次用了FFT来搞这种题目。
接下来还要训练几道这样的题目。
代码：
#include <iostream>#include <algorithm>#include <cstdio>#include <string>#include <cstring>#include <cmath>#include <vector>#include <list>#include <map>#include <set>#include <deque>#include <queue>#include <stack>#include <bitset>#include <functional>#include <sstream>#include <iomanip>#include <cmath>#include <cstdlib>#include <ctime>//#pragma comment(linker, "/STACK:102400000,102400000")typedef long long ll;#define INF 1e9const int maxn = 200005;//#define mod 1000000007#define eps 1e-7#define pi 3.1415926535897932384626433#define rep(i,n) for(int i=0;i<n;i++)#define rep1(i,n) for(int i=1;i<=n;i++)#define scan(n) scanf("%d",&n)#define scanll(n) scanf("%I64d",&n)#define scan2(n,m) scanf("%d%d",&n,&m)#define scans(s) scanf("%s",s);#define ini(a) memset(a,0,sizeof(a))#define out(n) printf("%d\n",n)//ll gcd(ll a,ll b) { return b==0?a:gcd(b,a%b);}using namespace std;#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1const double PI = acos(-1.0);//复数结构体struct complex{    double r,i;    complex(double _r = 0.0,double _i = 0.0)    {        r = _r; i = _i;    }    complex operator +(const complex &b)    {        return complex(r+b.r,i+b.i);    }    complex operator -(const complex &b)    {        return complex(r-b.r,i-b.i);    }    complex operator *(const complex &b)    {        return complex(r*b.r-i*b.i,r*b.i+i*b.r);    }};/* * 进行FFT和IFFT前的反转变换。 * 位置i和 （i二进制反转后位置）互换 * len必须去2的幂 */void change(complex 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]);        //交换互为小标反转的元素，i<j保证交换一次        //i做正常的+1，j左反转类型的+1,始终保持i和j是反转的        k = len/2;        while( j >= k)        {            j -= k;            k /= 2;        }        if(j < k) j += k;    }}/* * 做FFT * len必须为2^k形式， * on==1时是DFT，on==-1时是IDFT */void FFT(complex y[],int len,int on){    change(y,len);    for(int h = 2; h <= len; h <<= 1)    {        complex wn(cos(-on*2*PI/h),sin(-on*2*PI/h));        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(on == -1)        for(int i = 0;i < len;i++)            y[i].r /= len;}char a[maxn],b[maxn];complex A[maxn],B[maxn];int ans[maxn];int main(){#ifndef ONLINE_JUDGEfreopen("in.txt","r",stdin);//     freopen("out.txt","w",stdout);#endif  while(~scanf("%s%s",a,b)){int l1 = strlen(a), l2 = strlen(b);int n = 1;while(n < max(l1 , l2) * 2) n <<= 1; //n至少为maxl的两倍ini(A);ini(B);ini(ans);rep(i,n){if(i < l1) A[i] = complex(a[l1 - i - 1] - '0',0); //高位在前，这样FFT后高位在后else A[i] = complex(0,0);}rep(i,n){if(i < l2) B[i] = complex(b[l2 - i - 1] - '0',0);else B[i] = complex(0,0);}FFT(A,n,1);FFT(B,n,1);rep(i,n) A[i] = A[i] * B[i];FFT(A,n,-1);rep(i,n) ans[i] = (int)(A[i].r + 0.5);for(int i = 0;i < n; i++){ans[i+1] += ans[i] / 10; //向上进位ans[i] %= 10; //本身保留个位}for(; !ans[n]; n--); //消去前导0if(n < 0){ puts("0"); continue; }for(int i = n;i >= 0; i--)printf("%d",ans[i]);puts("");}return 0;}
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