就是求兩個長度爲50000的數相乘。
第一次用了FFT來搞這種題目。
接下來還要訓練幾道這樣的題目。
代碼:
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <string>
#include <cstring>
#include <cmath>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <functional>
#include <sstream>
#include <iomanip>
#include <cmath>
#include <cstdlib>
#include <ctime>
//#pragma comment(linker, "/STACK:102400000,102400000")
typedef long long ll;
#define INF 1e9
const int maxn = 200005;
//#define mod 1000000007
#define eps 1e-7
#define pi 3.1415926535897932384626433
#define rep(i,n) for(int i=0;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define scan(n) scanf("%d",&n)
#define scanll(n) scanf("%I64d",&n)
#define scan2(n,m) scanf("%d%d",&n,&m)
#define scans(s) scanf("%s",s);
#define ini(a) memset(a,0,sizeof(a))
#define out(n) printf("%d\n",n)
//ll gcd(ll a,ll b) { return b==0?a:gcd(b,a%b);}
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
const double PI = acos(-1.0);
//複數結構體
struct complex
{
double r,i;
complex(double _r = 0.0,double _i = 0.0)
{
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);
}
};
/*
* 進行FFT和IFFT前的反轉變換。
* 位置i和 (i二進制反轉後位置)互換
* len必須去2的冪
*/
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]);
//交換互爲小標反轉的元素,i<j保證交換一次
//i做正常的+1,j左反轉類型的+1,始終保持i和j是反轉的
k = len/2;
while( j >= k)
{
j -= k;
k /= 2;
}
if(j < k) j += k;
}
}
/*
* 做FFT
* len必須爲2^k形式,
* on==1時是DFT,on==-1時是IDFT
*/
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;
}
char a[maxn],b[maxn];
complex A[maxn],B[maxn];
int ans[maxn];
int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
// freopen("out.txt","w",stdout);
#endif
while(~scanf("%s%s",a,b))
{
int l1 = strlen(a), l2 = strlen(b);
int n = 1;
while(n < max(l1 , l2) * 2) n <<= 1; //n至少爲maxl的兩倍
ini(A);
ini(B);
ini(ans);
rep(i,n)
{
if(i < l1) A[i] = complex(a[l1 - i - 1] - '0',0); //高位在前,這樣FFT後高位在後
else A[i] = complex(0,0);
}
rep(i,n)
{
if(i < l2) B[i] = complex(b[l2 - i - 1] - '0',0);
else B[i] = complex(0,0);
}
FFT(A,n,1);
FFT(B,n,1);
rep(i,n) A[i] = A[i] * B[i];
FFT(A,n,-1);
rep(i,n) ans[i] = (int)(A[i].r + 0.5);
for(int i = 0;i < n; i++)
{
ans[i+1] += ans[i] / 10; //向上進位
ans[i] %= 10; //本身保留個位
}
for(; !ans[n]; n--); //消去前導0
if(n < 0){ puts("0"); continue; }
for(int i = n;i >= 0; i--)
printf("%d",ans[i]);
puts("");
}
return 0;
}