首页 > 代码库 > HDU 1402 A * B Problem Plus FFT
HDU 1402 A * B Problem Plus FFT
A * B Problem Plus
Problem Description
Calculate A * B.
Input
Each line will contain two integers A and B. Process to end of file.
Note: the length of each integer will not exceed 50000.
Note: the length of each integer will not exceed 50000.
Output
For each case, output A * B in one line.
Sample Input
1210002
Sample Output
22000
题解:
FFT入门
注意几组数据
0 0
0 5
0005 000006
#include<bits/stdc++.h>using namespace std;#pragma comment(linker, "/STACK:102400000,102400000")#define ls i<<1#define rs ls | 1#define mid ((ll+rr)>>1)#define pii pair<int,int>#define MP make_pairtypedef long long LL;const long long INF = 1e18+1LL;const double pi = acos(-1.0);const int N = 1e5+10, M = 1e3+20,inf = 2e9,mod = 1e9+7;struct Complex { double r , i ; Complex () {} Complex ( double r , double i ) : r ( r ) , i ( i ) {} Complex operator + ( const Complex& t ) const { return Complex ( r + t.r , i + t.i ) ; } Complex operator - ( const Complex& t ) const { return Complex ( r - t.r , i - t.i ) ; } Complex operator * ( const Complex& t ) const { return Complex ( r * t.r - i * t.i , r * t.i + i * t.r ) ; }} ;void FFT ( Complex y[] , int n , int rev ) { for ( int i = 1 , j , t , k ; i < n ; ++ i ) { for ( j = 0 , t = i , k = n >> 1 ; k ; k >>= 1 , t >>= 1 ) j = j << 1 | t & 1 ; if ( i < j ) swap ( y[i] , y[j] ) ; } for ( int s = 2 , ds = 1 ; s <= n ; ds = s , s <<= 1 ) { Complex wn = Complex ( cos ( rev * 2 * pi / s ) , sin ( rev * 2 * pi / s ) ) , w ( 1 , 0 ) , t ; for ( int k = 0 ; k < ds ; ++ k , w = w * wn ) { for ( int i = k ; i < n ; i += s ) { y[i + ds] = y[i] - ( t = w * y[i + ds] ) ; y[i] = y[i] + t ; } } } if ( rev == -1 ) for ( int i = 0 ; i < n ; ++ i ) y[i].r /= n ;}Complex s[N*4],t[N*4];char a[N],b[N];int ans[N];int main() { while(scanf("%s%s",a,b)!=EOF) { int n = strlen(a); int m = strlen(b); int flag2 = 0, flag1 = 0; for(int i = 0; i < n; ++i) { if(a[i] == ‘0‘ && !flag1) { continue; } else a[flag1++] = a[i]; } for(int i = 0; i < m; ++i) { if(b[i] == ‘0‘ && !flag2) { continue; } else b[flag2++] = b[i]; } n = flag1; m = flag2; // cout<<n<<" "<<m<<endl; if(n == 0 || m == 0) { puts("0"); continue; } int n1 = 1; while(n1 <= m+n-2) n1<<=1; for(int i = 0; i < n; ++i) { s[i] = Complex(a[i] - ‘0‘,0); } for(int i = n; i < n1; ++i) { s[i] = Complex(0,0); } for(int i = 0; i < m; ++i) { t[i] = Complex(b[i] - ‘0‘,0); } for(int i = m; i < n1; ++i) { t[i] = Complex(0,0); } FFT(s,n1,1); FFT(t,n1,1); for(int i = 0; i < n1; ++i) s[i] = s[i]*t[i]; FFT(s,n1,-1); int f = 1; int last = 0,cnt = 0; for(int i = n+m-2; i >= 0; --i) { int x = (int) (s[i].r+0.1) + last; //printf("%d\n",x); last = 0; if(x > 9) { last+=x/10; ans[++cnt] = x % 10; } else ans[++cnt] = x; } if(last) { ans[++cnt] = last; } for(int i = cnt; i >= 1; --i) printf("%d",ans[i]); printf("\n"); } return 0;}
HDU 1402 A * B Problem Plus FFT
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。