http://acm.hdu.edu.cn/showproblem.php?pid=1402
我的代码，由于时间的问题，以后找机会优化了！（超时）
View Code
#include " iostream "
#define M 100010
using namespace std;
char ch1[M],ch2[M];
int a[M],b[M];
int c[M],d[M];
int num[M];
int H[M];
int t,k,g,v;
int i,j;
int La,Lb;
int sign = 0 ;
void Add( int x , int y)
{
sign = 0 ; t = 0 ;
int p,q,flag = 0 ;
if (x > y)
{
for (p = 0 ,q = 0 ;p < y,q < y; p ++ ,q ++ )
{
num[t ++ ] = (d[p] + H[q] + flag) % 10 ;
flag = (d[p] + H[q] + flag) % 10 ;
}
for (p = y; p < x ; p ++ )
{
num[t ++ ] = (d[p] + flag) % 10 ;
flag = (d[p] + flag) / 10 ;
}
}
else if (x == y)
{
for (p = 0 ,q = 0 ;p < x,q < y; p ++ ,q ++ )
{
num[t ++ ] = (d[p] + H[q] + flag) % 10 ;
flag = (d[p] + H[q] + flag) / 10 ;
}
}
else
{
flag = 0 ;
for (p = 0 ,q = 0 ;p < x,q < x; p ++ ,q ++ )
{
num[t ++ ] = (H[p] + d[q] + flag) % 10 ;
flag = (H[p] + d[q] + flag) / 10 ;
}
for (q = x;q < y;q ++ )
{
num[t ++ ] = (H[q] + flag) % 10 ;
flag = (H[q] + flag) / 10 ;
}
}
while (flag)
{
num[t ++ ] = flag % 10 ;
flag /= 10 ;
}
for ( int yy = 0 ; yy < t;yy ++ )
{
d[yy] = num[yy];
if (d[yy] == 0 ) sign ++ ; // sign用于标记都为结果0的情况
}
g = t;
memset(H , 0 , sizeof (H));
memset(num, 0 , sizeof (num));
}
void BigN()
{
int flag = 0 ;
int mark = 0 ;
if (La > Lb)
{
for (i = Lb - 1 ;i >= 0 ;i -- )
{
k = 0 ;
for (j = La - 1 ;j >= 0 ;j -- )
{
c[k ++ ] = (b[i] * a[j] + flag) % 10 ;
flag = (b[i] * a[j] + flag) / 10 ;
}
while (flag)
{
c[k ++ ] = flag % 10 ;
flag /= 10 ;
}
// 末尾赋值0
v = 0 ;
for ( int xx = 0 ; xx < mark; xx ++ )
H[v ++ ] = 0 ;
for ( int m = 0 ;m < k;m ++ )
H[v ++ ] = c[m];
memset(c, 0 , sizeof (c));
// ****************************
Add(g , v);
mark ++ ;
}
}
else
{
for (i = La - 1 ;i >= 0 ;i -- )
{
k = 0 ;
for (j = Lb - 1 ;j >= 0 ;j -- )
{
c[k ++ ] = (a[i] * b[j] + flag) % 10 ;
flag = (a[i] * b[j] + flag) / 10 ;
}
while (flag)
{
c[k ++ ] = flag % 10 ;
flag /= 10 ;
}
// 末尾赋值0
v = 0 ;
for ( int xx = 0 ; xx < mark; xx ++ )
H[v ++ ] = 0 ;
for ( int m = 0 ;m < k;m ++ )
H[v ++ ] = c[m];
// ****************************
Add(g , v);
mark ++ ;
}
}
}
int main()
{
while (cin >> ch1 >> ch2)
{
t = 0 ;k = 0 ;g = 0 ;
La = strlen(ch1);
Lb = strlen(ch2);
for (i = 0 ;i < La;i ++ ) a[i] = ch1[i] - ' 0 ' ;
for (i = 0 ;i < Lb;i ++ ) b[i] = ch2[i] - ' 0 ' ;
BigN();
if (sign != t)
{
for (i = t - 1 ;i >= 0 ;i -- )
cout << d[i];
cout << endl;
}
else cout << " 0 " << endl;
}
return 0 ;
}
一位dn的，神人啊！
http://www.cppblog.com/misschuer/archive/2010/01/16/80356.html
View Code
#include < iostream >
#include < cmath >
using namespace std;
typedef struct vir{
double re,im;
vir(){}
vir( double a, double b){re = a;im = b;}
vir operator + ( const vir & b){ return vir(re + b.re,im + b.im);}
vir operator - ( const vir & b){ return vir(re - b.re,im - b.im);}
vir operator * ( const vir & b){ return vir(re * b.re - im * b.im,re * b.im + b.re * im);}
}vir;
vir x1[ 200005 ],x2[ 200005 ];
const double Pi = acos( - 1.0 );
void change(vir * x, int len, int loglen)
{
int i,j,k,t;
for (i = 0 ;i < len;i ++ )
{
t = i;
for (j = k = 0 ;j < loglen;j ++ ,t >>= 1 )
k = (k << 1 ) | (t & 1 );
if (k < i)
{
vir wt = x[k];
x[k] = x[i];
x[i] = wt;
}
}
}
void fft(vir * x, int len, int loglen)
{
int i,j,t,s,e;
change(x,len,loglen);
t = 1 ;
for (i = 0 ;i < loglen;i ++ ,t <<= 1 )
{
s = 0 ;
e = s + t;
while (s < len)
{
vir a,b,wo(cos(Pi / t),sin(Pi / t)),wn( 1 , 0 );
for (j = s;j < s + t;j ++ )
{
a = x[j];
b = x[j + t] * wn;
x[j] = a + b;
x[j + t] = a - b;
wn = wn * wo;
}
s = e + t;
e = s + t;
}
}
}
void dit_fft(vir * x, int len, int loglen)
{
int i,j,s,e,t = 1 << loglen;
for (i = 0 ;i < loglen;i ++ )
{
t >>= 1 ;
s = 0 ;
e = s + t;
while (s < len)
{
vir a,b,wn( 1 , 0 ),wo(cos(Pi / t), - sin(Pi / t));
for (j = s;j < s + t;j ++ )
{
a = x[j] + x[j + t];
b = (x[j] - x[j + t]) * wn;
x[j] = a;
x[j + t] = b;
wn = wn * wo;
}
s = e + t;
e = s + t;
}
}
change(x,len,loglen);
for (i = 0 ;i < len;i ++ )
x[i].re /= len;
}
int main()
{
char a[ 100005 ],b[ 100005 ];
int i,len1,len2,t,over,len,loglen;
while (scanf( " %s%s " ,a,b) != EOF)
{
len1 = strlen(a) << 1 ;
len2 = strlen(b) << 1 ;
len = 1 ;
loglen = 0 ;
while (len < len1)
{
len <<= 1 ;
loglen ++ ;
}
while (len < len2)
{
len <<= 1 ;
loglen ++ ;
}
for (i = 0 ;a[i] != ' \0 ' ;i ++ )
{
x1[i].re = a[i] - ' 0 ' ;
x1[i].im = 0 ;
}
for (;i < len;i ++ )
x1[i].re = x1[i].im = 0 ;
for (i = 0 ;b[i] != ' \0 ' ;i ++ )
{
x2[i].re = b[i] - ' 0 ' ;
x2[i].im = 0 ;
}
for (;i < len;i ++ )
x2[i].re = x2[i].im = 0 ;
fft(x1,len,loglen);
fft(x2,len,loglen);
for (i = 0 ;i < len;i ++ )
x1[i] = x1[i] * x2[i];
dit_fft(x1,len,loglen);
for (i = (len1 + len2) / 2 - 2 ,over = loglen = 0 ;i >= 0 ;i -- )
{
t = x1[i].re + over + 0.5 ;
a[loglen ++ ] = t % 10 ;
over = t / 10 ;
}
while (over)
{
a[loglen ++ ] = over % 10 ;
over /= 10 ;
}
for (loglen -- ;loglen >= 0 &&! a[loglen];loglen -- );
if (loglen < 0 )
putchar( ' 0 ' );
else
for (;loglen >= 0 ;loglen -- )
putchar(a[loglen] + ' 0 ' );
putchar( ' \n ' );
}
return 0 ;
}
转载于:https://my.oschina.net/garyun/blog/602828