你最愿意做的哪件事,才是你的天赋所在

0%

HDU1402(FFT)

题目链接

HDU1402

思路

FFT板子题,注意前置零的判断以及数组大小,
不会FFT的戳
FFT详解

代码实现

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#include<string>
#include<cstdio>
#include<math.h>
#include<iostream>
#include<string.h>
#include <algorithm>
using namespace std;
#define ll long long
const double pi = acos(-1);
const int N = 3e5+10;
char s1[50010],s2[50010];
int a[N],b[N],r[N],res[N];
struct cp
{
double r,i;
cp(double _r = 0,double _i = 0)
{
r = _r; i = _i;
}
cp operator +(const cp &b)
{
return cp(r+b.r,i+b.i);
}
cp operator -(const cp &b)
{
return cp(r-b.r,i-b.i);
}
cp operator *(const cp &b)
{
return cp(r*b.r-i*b.i,r*b.i+i*b.r);
}
}A[N],B[N];
void FFT(cp *s,int n,int inv){
for(int i=0;i<n;i++){
if(i<r[i])swap(s[i],s[r[i]]);
}
for(int len=2;len<=n;len<<=1){
cp wn = cp(cos(2*pi/len),inv*sin(2*pi/len));
for(int st=0;st<n;st+=len){
cp w = cp(1.0,0.0);
for(int i=st;i<st+len/2;i++,w=w*wn){
cp x=s[i];
cp y=w*s[i+len/2];
s[i]=x+y;
s[i+len/2]=x-y;
}
}
}
if(inv==-1){
for(int i=0;i<n;i++)
s[i].r/=n;
}
}
void solve(){
int len =1,bit=0;
int s1len=strlen(s1);
int s2len=strlen(s2);
int slen = max(s1len,s2len);
while(len<slen*2)len<<=1,bit++;
for(int i=0;i<len;i++){
a[i]=b[i]=r[i]=res[i]=0;
A[i]=B[i]=cp(0.0,0.0);
}
for(int i=0;i<s1len;i++)a[i]=s1[i]-'0',A[i].r=a[i];
for(int i=0;i<s2len;i++)b[i]=s2[i]-'0',B[i].r=b[i];
for(int i=0;i<len;i++)r[i]=(r[i>>1]>>1)|((i&1)<<(bit-1));
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);
for(int i=0;i<len;i++)res[i]=(int)(A[i].r+0.5);
for(int i=len-1;i>=1;i--){
if(res[i]>=10)res[i-1]+=res[i]/10,res[i]%=10;
}
if(res[0]==0)puts("0");
else {
for(int i=0;i<s1len+s2len-1;i++)
{
printf("%d",res[i]);
}
printf("\n");
}

}
int main(){
while(scanf("%s %s",s1,s2)!=EOF){
solve();
}
return 0;
}
-------------你最愿意做的哪件事才是你的天赋所在-------------