FFT A*B模板

从卿学姐那里偷了份FFT大数相乘的模板。。。果断交了一发HDU 1402

还不孬

  1 //FFT 大整数乘法
  2 #include<cstdio>
  3 #include<cmath>
  4 #include<cstring>
  5 #include<algorithm>
  6 
  7 using namespace std;
  8 
  9 
 10 const int N = 500005;
 11 const double pi = acos(-1.0);
 12 
 13 char s1[N],s2[N];
 14 int len,res[N];
 15 
 16 struct Complex
 17 {
 18     double r,i;
 19     Complex(double r=0,double i=0):r(r),i(i) {};
 20     Complex operator+(const Complex &rhs)
 21     {
 22         return Complex(r + rhs.r,i + rhs.i);
 23     }
 24     Complex operator-(const Complex &rhs)
 25     {
 26         return Complex(r - rhs.r,i - rhs.i);
 27     }
 28     Complex operator*(const Complex &rhs)
 29     {
 30         return Complex(r*rhs.r - i*rhs.i,i*rhs.r + r*rhs.i);
 31     }
 32 } va[N],vb[N];
 33 
 34 void rader(Complex F[],int len) //len = 2^M,reverse F[i] with  F[j] j为i二进制反转
 35 {
 36     int j = len >> 1;
 37     for(int i = 1;i < len - 1;++i)
 38     {
 39         if(i < j) swap(F[i],F[j]);  // reverse
 40         int k = len>>1;
 41         while(j>=k)
 42         {
 43             j -= k;
 44             k >>= 1;
 45         }
 46         if(j < k) j += k;
 47     }
 48 }
 49 
 50 void FFT(Complex F[],int len,int t)
 51 {
 52     rader(F,len);
 53     for(int h=2;h<=len;h<<=1)
 54     {
 55         Complex wn(cos(-t*2*pi/h),sin(-t*2*pi/h));
 56         for(int j=0;j<len;j+=h)
 57         {
 58             Complex E(1,0); //旋转因子
 59             for(int k=j;k<j+h/2;++k)
 60             {
 61                 Complex u = F[k];
 62                 Complex v = E*F[k+h/2];
 63                 F[k] = u+v;
 64                 F[k+h/2] = u-v;
 65                 E=E*wn;
 66             }
 67         }
 68     }
 69     if(t==-1)   //IDFT
 70         for(int i=0;i<len;++i)
 71             F[i].r/=len;
 72 }
 73 
 74 void Conv(Complex a[],Complex b[],int len) //求卷积
 75 {
 76     FFT(a,len,1);
 77     FFT(b,len,1);
 78     for(int i=0;i<len;++i) a[i] = a[i]*b[i];
 79     FFT(a,len,-1);
 80 }
 81 
 82 void init(char *s1,char *s2)
 83 {
 84     int n1 = strlen(s1),n2 = strlen(s2);
 85     len = 1;
 86     while(len < 2*n1 || len < 2*n2) len <<= 1;
 87     int i;
 88     for(i=0;i<n1;++i)
 89     {
 90         va[i].r = s1[n1-i-1]-'0';
 91         va[i].i = 0;
 92     }
 93     while(i<len)
 94     {
 95         va[i].r = va[i].i = 0;
 96         ++i;
 97     }
 98     for(i=0;i<n2;++i)
 99     {
100         vb[i].r = s2[n2-i-1]-'0';
101         vb[i].i = 0;
102     }
103     while(i<len)
104     {
105         vb[i].r = vb[i].i = 0;
106         ++i;
107     }
108 }
109 
110 void gao()
111 {
112     Conv(va,vb,len);
113     memset(res,0,sizeof res);
114     for(int i=0;i<len;++i)
115     {
116         res[i]=va[i].r + 0.5;
117     }
118     for(int i=0;i<len;++i)
119     {
120         res[i+1]+=res[i]/10;
121         res[i]%=10;
122     }
123     int high = 0;
124     for(int i=len-1;i>=0;--i)
125     {
126         if(res[i])
127         {
128             high = i;
129             break;
130         }
131     }
132     for(int i=high;i>=0;--i) putchar('0'+res[i]);
133     puts("");
134 }
135 
136 
137 int main()
138 {
139     while(scanf("%s%s",s1,s2)==2)
140     {
141         init(s1,s2);
142         gao();
143     }
144     return 0;
145 }

 

posted on 2016-09-29 13:20  disppr  阅读(533)  评论(0)  编辑  收藏  举报