日本久久久精品中文字幕,欧美亚洲国产精品久久,日韩欧美亚洲综合久久影院d3

hdu 1402 A * B Problem Plus

http://acm.hdu.edu.cn/showproblem.php?pid=1402

#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;

}

posted on 2009-04-18 15:59 此最相思閱讀(1893) 評論(6) 編輯收藏引用

# re: hdu 1402 A * B Problem Plus 2009-10-13 20:18 zhou

# re: hdu 1402 A * B Problem Plus 2009-10-13 20:20 啊是

# re: hdu 1402 A * B Problem Plus 2009-10-13 20:35 zhou

# re: hdu 1402 A * B Problem Plus 2009-10-13 20:36 啊是

# re: hdu 1402 A * B Problem Plus 2010-01-16 00:43 abilitytao

# re: hdu 1402 A * B Problem Plus 2011-04-03 23:13 xiaomai

misschuer

導(dǎo)航

公告

留言簿

隨筆分類

文章分類

隨筆檔案

文章檔案

閱讀排行榜

評論排行榜

常用鏈接

統(tǒng)計

積分與排名

百事通

最新評論

hdu 1402 A * B Problem Plus

評論

只有注冊用戶登錄后才能發(fā)表評論。
【推薦】100%開源！大型工業(yè)跨平臺軟件C++源碼提供，建模，組態(tài)！



網(wǎng)站導(dǎo)航: 博客園 IT新聞 BlogJava 博問 Chat2DB 管理