HDU 1402 A * B Problem Plus ——（大数乘法，FFT）

　　因为刚学fft，想拿这题练练手，结果WA了个爽= =。

　　总结几点犯的错误：

　　1.要注意处理前导零的问题。

　　2.一定要注意数组大小的问题。（前一个fft的题因为没用到b数组，所以b就没管，这里使用了b数组，结果忘记给其大小乘以4倍了）

　　代码如下：

#include<bits/stdc++.h>
using namespace std;
const double pi = atan(1.0)*4;
const int N = 5e4 + 5;
typedef long long ll;

struct Complex {
    double x,y;
    Complex(double _x=0,double _y=0)
        :x(_x),y(_y) {}
    Complex operator + (Complex &tt) { return Complex(x+tt.x,y+tt.y); }
    Complex operator - (Complex &tt) { return Complex(x-tt.x,y-tt.y); }
    Complex operator * (Complex &tt) { return Complex(x*tt.x-y*tt.y,x*tt.y+y*tt.x); }
};
Complex a[N*4],b[N*4];
void fft(Complex *a, int n, int rev) {
    // n是(大于等于相乘的两个数组长度)2的幂次 ； 比如长度是5 ,那么 n = 8    2^2 < 5          2^3 > 5
    // 从0开始表示长度，对a进行操作
    // rev==1进行DFT，==-1进行IDFT
    for (int i = 1,j = 0; i < n; ++ i) {
        for (int k = n>>1; k > (j^=k); k >>= 1);
        if (i<j) std::swap(a[i],a[j]);
    }
    for (int m = 2; m <= n; m <<= 1) {
        Complex wm(cos(2*pi*rev/m),sin(2*pi*rev/m));
        for (int i = 0; i < n; i += m) {
            Complex w(1.0,0.0);
            for (int j = i; j < i+m/2; ++ j) {
                Complex t = w*a[j+m/2];
                a[j+m/2] = a[j] - t;
                a[j] = a[j] + t;
                w = w * wm;
            }
        }
    }
    if (rev==-1) {
        for (int i = 0; i < n; ++ i) a[i].x /= n,a[i].y /= n;
    }
}

char s[N], t[N];
int c[N*2];

int main(){
    /*a[0] = Complex(0,0); //  a[0]: x的0次项。
    a[1] = Complex(1,0);
    a[2] = Complex(2,0);
    a[3] = Complex(3,0);

    b[0] = Complex(3,0);
    b[1] = Complex(2,0);
    b[2] = Complex(1,0);
    b[3] = Complex(0,0);
    fft(a,8,1);
    fft(b,8,1);
    for(int i = 0 ; i < 8 ; i ++){
        a[i] = a[i] * b[i];
    }
    fft(a,8,-1);
    for(int i = 0 ; i < 8 ; i ++){
        cout << i << "   " << a[i].x << endl;;
    }*/
    while(scanf("%s%s",s,t) == 2)
    {
        int n = strlen(s), m = strlen(t);
        for(int i=0;i<n;i++) a[i] = Complex(s[n-i-1]-'0', 0);
        for(int i=0;i<m;i++) b[i] = Complex(t[m-i-1]-'0', 0);
        int len = n+m-1;
        int LIM = 1;
        while(LIM < len) LIM <<= 1;
        for(int i=n;i<LIM;i++) a[i] = Complex(0, 0);
        for(int i=m;i<LIM;i++) b[i] = Complex(0, 0);
        fft(a, LIM, 1);
        fft(b, LIM, 1);
        for(int i=0;i<LIM;i++) a[i] = a[i] * b[i];
        fft(a, LIM, -1);
        memset(c, 0, sizeof c);
        for(int i=0;i<len-1;i++)
        {
            c[i] += (int)(a[i].x + 0.1);
            c[i+1] += c[i] / 10;
            c[i] %= 10;
        }
        c[len-1] += (int)(a[len-1].x + 0.1);
        if(c[len-1] > 9)
        {
            c[len] = c[len-1] / 10;
            c[len-1] %= 10;
            len++;
        }
        for(int i=len-1;i>0;i--) if(c[i] == 0) len--; else break; // 0 * 345 = 0 instead of 000
        for(int i=0;i<len/2;i++) swap(c[i], c[len-i-1]);
        for(int i=0;i<len;i++) printf("%d",c[i]);
        puts("");
    }
    return 0;
}