数论-FFT高精度乘法

NKOJ3071 模板题：求两个整数之积.

FFT 函数里 ty = 1 表示 DFT 运算，ty = -1 表示 IDFT 运算.

 1 #include <stdio.h>
 2 #include <complex>
 3 
 4 using namespace std;
 5 
 6 typedef complex<double> CP;
 7 typedef long long LL;
 8 
 9 const int _N = 300005;
10 const double PI = 3.1415926535897932384626433832795;
11 
12 LL rev[_N], Ans[_N];
13 char str1[_N], str2[_N];
14 CP X[_N], Y[_N];
15 
16 void GetRev(LL bit)
17 {
18     for (LL i = 0; i < (1<<bit); ++i)
19         rev[i] = (rev[i>>1]>>1) | ((i&1)<<(bit-1));
20     return;
21 }
22 
23 void FFT(CP *A, LL n, LL ty)
24 {
25     LL i, k, len;
26     for (i = 0; i < n; ++i)
27         if (i < rev[i]) swap(A[i], A[rev[i]]);
28         
29     for (len = 1; len < n; len <<= 1) {
30         CP wn = exp(CP(0, ty*PI/len));
31         for (i = 0; i < n; i += len<<1) {
32             CP wi(1, 0);
33             for (k = i; k < i+len; ++k) {
34                 CP t0 = A[k], t1 = wi*A[k+len];
35                 A[k] = t0+t1, A[k+len] = t0-t1;
36                 wi *= wn;
37             }
38         }
39     }
40     if (ty == -1)
41         for (i = 0; i < n; ++i) A[i] /= n;
42     return;
43 }
44 
45 bool Input(LL &len, char *str)
46 {
47     char tt; bool flag = false; int i;
48     while (((tt = getchar()) < '0' || tt > '9') && tt != '-');
49     if (tt == '-') flag = true, i = -1;
50     else str[0] = tt, i = 0;
51     while ((tt = getchar()) >= '0' && tt <= '9') str[++i] = tt;
52     len = i+1;
53     return flag;
54 }
55 
56 int main()
57 {
58     LL flag1, flag2, l1, l2, i;
59     flag1 = Input(l1, str1), flag2 = Input(l2, str2);
60 //    printf("

%s %s

 %lld %lld
", str1, str2, l1, l2);
61     LL a = 1, x = 0;
62     while (a < l1+l2-1) a <<= 1, ++x;
63 //    printf("

a = %lld, x = %lld

", a, x);
64     for (i = 0; i < l1; ++i) X[i] = (double)(str1[l1-1-i]-'0');
65     for (i = 0; i < l2; ++i) Y[i] = (double)(str2[l2-1-i]-'0');
66     GetRev(x), FFT(X, a, 1), FFT(Y, a, 1);
67     for (i = 0; i < a; ++i) X[i] *= Y[i];
68     FFT(X, a, -1);
69     for (i = 0; i < l1+l2; ++i)
70         Ans[i] += (long long)(X[i].real() + 0.5), Ans[i+1] += Ans[i]/10, Ans[i] %= 10;
71     for (i = l1+l2; i >= 0 && !Ans[i]; --i);
72     if (i == -1) { printf("0
"); return 0; }
73     if (flag1 ^ flag2) putchar('-');
74     while (i >= 0) putchar(Ans[i] + '0'), --i;
75     putchar('
');
76     return 0;
77 }

题目

P3071【高精度】a*b

时间限制 : 10000 MS 空间限制 : 65536 KB

问题描述

给你两个正整数a,b，计算它们的乘积。

输入格式

第一行一个正整数a
第二行一个正整数b

输出格式

一行，表示a*b

样例输入

111222333444555666777888999
999888777666555444333222111

样例输出

111209963037098814851876554444456814851901296370456889

提示

a,b分别不超过100000位

来源感谢nodgd放题并提供数据