HDU 1402 A * B Problem Plus FFT

A * B Problem Plus

[align=left][b]Problem Description[/b][/align]
Calculate A * B.

[align=left][b]Input[/b][/align]
Each line will contain two integers A and B. Process to end of file.

Note: the length of each integer will not exceed 50000.

[align=left][b]Output[/b][/align]
For each case, output A * B in one line.

[align=left][b]Sample Input[/b][/align]

1
2
1000
2

[align=left][b]Sample Output[/b][/align]

2
2000

[align=left][b]题解：[/b][/align]
[align=left]　　FFT入门[/align]
[align=left]　　注意几组数据[/align]
[align=left]　　0 0[/align]
[align=left]　　0 5[/align]
[align=left] 0005 000006[/align]

#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_pair
typedef 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
,b
;
int ans
;
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;
}

[align=left]　　[/align]