本文主要是介绍hdu 1402 A * B Problem Plus[【FFT】,希望对大家解决编程问题提供一定的参考价值,需要的开发者们随着小编来一起学习吧!
这是一道FFT模板题,然而菜鸟还是WA好多发。。。
代码:
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<map>
#include<string>
#include<queue>
#include<vector>
#include<list>
#include<bitset>
//#pragma comment(linker,"/STACK:1024000000,1024000000")
using namespace std;
typedef long long ll;
#define INF 0x3f3f3f3f
#define N 200005
const double PI = acos(-1.0);
struct complex
{double r,i;complex(double _r = 0,double _i = 0){r = _r; i = _i;}complex operator +(const complex &b){return complex(r+b.r,i+b.i);}complex operator -(const complex &b){return complex(r-b.r,i-b.i);}complex operator *(const complex &b){return complex(r*b.r-i*b.i,r*b.i+i*b.r);}
};
void change(complex y[],int len)
{int i,j,k;for(i = 1, j = len/2;i < len-1;i++){if(i < j)swap(y[i],y[j]);k = len/2;while( j >= k){j -= k;k /= 2;}if(j < k)j += k;}
}
void fft(complex y[],int len,int on)
{change(y,len);for(int h = 2;h <= len;h <<= 1){complex wn(cos(-on*2*PI/h),sin(-on*2*PI/h));for(int j = 0;j < len;j += h){complex w(1,0);for(int k = j;k < j+h/2;k++){complex u = y[k];complex t = w*y[k+h/2];y[k] = u+t;y[k+h/2] = u-t;w = w*wn;}}}if(on == -1)for(int i = 0;i < len;i++)y[i].r /= len;
}
complex x1[N],x2[N];
char a[N/4],b[N/4];
int num[N];
int main()
{while(~scanf("%s%s",a,b)){int len1=strlen(a);int len2=strlen(b);int l=1;while(l<len1+len2) l<<=1;for(int i=len1-1;i>=0;i--) x1[i]=complex(a[len1-i-1]-'0',0);for(int i=len1;i<l;i++) x1[i]=complex(0,0);for(int i=len2-1;i>=0;i--) x2[i]=complex(b[len2-i-1]-'0',0);for(int i=len2;i<l;i++) x2[i]=complex(0,0);fft(x1,l,1);fft(x2,l,1);for(int i=0;i<l;i++) x1[i]=x1[i]*x2[i];fft(x1,l,-1);for(int i=0;i<l;i++) num[i]=x1[i].r+0.5;for(int i=0;i<l;i++){num[i+1]+=num[i]/10;num[i]%=10;}l=len1+len2-1;int f=0;for(int i=l;i>=0;i--){if(!f&&num[i]==0) continue;f=1;printf("%d",num[i]);}if(!f) puts("0");else puts("");}return 0;
}
这篇关于hdu 1402 A * B Problem Plus[【FFT】的文章就介绍到这儿,希望我们推荐的文章对编程师们有所帮助!