HDU 1402 A * B Problem Plus (FFT模板题)
2016-10-07 12:45
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A * B Problem Plus
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 18229 Accepted Submission(s): 4080
[align=left]Problem Description[/align]
Calculate A * B.
[align=left]Input[/align]
Each line will contain two integers A and B. Process to end of file.
Note: the length of each integer will not exceed 50000.
[align=left]Output[/align]
For each case, output A * B in one line.
[align=left]Sample Input[/align]
1
2
1000
2
[align=left]Sample Output[/align]
2
2000
[align=left]Author[/align]
DOOM III
[align=left]Recommend[/align]
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题解:FFT模板题。大整数乘法。
AC代码:
//FFT 大整数乘法 #include<cstdio> #include<cmath> #include<cstring> #include<algorithm> using namespace std; const int N = 500005; const double pi = acos(-1.0); char s1 ,s2 ; int len,res ; struct Complex { double r,i; Complex(double r=0,double i=0):r(r),i(i) {}; Complex operator+(const Complex &rhs) { return Complex(r + rhs.r,i + rhs.i); } Complex operator-(const Complex &rhs) { return Complex(r - rhs.r,i - rhs.i); } Complex operator*(const Complex &rhs) { return Complex(r*rhs.r - i*rhs.i,i*rhs.r + r*rhs.i); } } va ,vb ; void rader(Complex F[],int len) //len = 2^M,reverse F[i] with F[j] j为i二进制反转 { int j = len >> 1; for(int i = 1;i < len - 1;++i) { if(i < j) swap(F[i],F[j]); // reverse int k = len>>1; while(j>=k) { j -= k; k >>= 1; } if(j < k) j += k; } } void FFT(Complex F[],int len,int t) { rader(F,len); for(int h=2;h<=len;h<<=1) { Complex wn(cos(-t*2*pi/h),sin(-t*2*pi/h)); for(int j=0;j<len;j+=h) { Complex E(1,0); //旋转因子 for(int k=j;k<j+h/2;++k) { Complex u = F[k]; Complex v = E*F[k+h/2]; F[k] = u+v; F[k+h/2] = u-v; E=E*wn; } } } if(t==-1) //IDFT for(int i=0;i<len;++i) F[i].r/=len; } void Conv(Complex a[],Complex b[],int len) //求卷积 { FFT(a,len,1); FFT(b,len,1); for(int i=0;i<len;++i) a[i] = a[i]*b[i]; FFT(a,len,-1); } void init(char *s1,char *s2) { int n1 = strlen(s1),n2 = strlen(s2); len = 1; while(len < 2*n1 || len < 2*n2) len <<= 1; int i; for(i=0;i<n1;++i) { va[i].r = s1[n1-i-1]-'0'; va[i].i = 0; } while(i<len) { va[i].r = va[i].i = 0; ++i; } for(i=0;i<n2;++i) { vb[i].r = s2[n2-i-1]-'0'; vb[i].i = 0; } while(i<len) { vb[i].r = vb[i].i = 0; ++i; } } void gao() { Conv(va,vb,len); memset(res,0,sizeof res); for(int i=0;i<len;++i) { res[i]=va[i].r + 0.5; } for(int i=0;i<len;++i) { res[i+1]+=res[i]/10; res[i]%=10; } int high = 0; for(int i=len-1;i>=0;--i) { if(res[i]) { high = i; break; } } for(int i=high;i>=0;--i) putchar('0'+res[i]); puts(""); } int main() { while(scanf("%s %s",s1,s2)==2) { init(s1,s2); gao(); } return 0; }
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