SWUST OJ 2540 少女输出问题

题目链接：见这里
题意: 中文题目， FFT裸题。
代码如下：

////Created by BLUEBUFF 2016/1/11//Copyright (c) 2016 BLUEBUFF.All Rights Reserved//#pragma comment(linker,"/STACK:102400000,102400000")//#include <ext/pb_ds/assoc_container.hpp>//#include <ext/pb_ds/tree_policy.hpp>//#include <ext/pb_ds/hash_policy.hpp>//#include <bits/stdc++.h>#include <set>#include <map>#include <queue>#include <stack>#include <cmath>#include <cstdio>#include <time.h>#include <cstdlib>#include <cstring>#include <complex>#include <sstream> //isstringstream#include <iostream>#include <algorithm>using namespace std;//using namespace __gnu_pbds;typedef long long LL;typedef unsigned long long uLL;typedef pair<int, LL> pp;#define REP1(i, a, b) for(int i = a; i < b; i++)#define REP2(i, a, b) for(int i = a; i <= b; i++)#define REP3(i, a, b) for(int i = a; i >= b; i--)#define CLR(a, b)     memset(a, b, sizeof(a))#define MP(x, y)      make_pair(x,y)template <class T1, class T2>inline void getmax(T1 &a, T2 b) { if (b>a)a = b; }template <class T1, class T2>inline void getmin(T1 &a, T2 b) { if (b<a)a = b; }const int maxn = 270010;const int maxm = 1e5+5;const int maxs = 10;const int maxp = 1e3 + 10;const int INF  = 1e9;const int UNF  = -1e9;const int mod  = 1e9 + 7;const int rev = (mod + 1) >> 1; // FWTconst double PI = acos(-1);//headstruct FastIO{    static const int S = 1310720;    int wpos;    char wbuf[S];    FastIO() : wpos(0) {}    inline int xchar()    {        static char buf[S];        static int len = 0, pos = 0;        if (pos == len)            pos = 0, len = fread(buf, 1, S, stdin);        if (pos == len) return -1;        return buf[pos ++];    }    inline int xuint()    {        int c = xchar(), x = 0;        while (c <= 32) c = xchar();        for (; '0' <= c && c <= '9'; c = xchar()) x = x * 10 + c - '0';        return x;    }    inline int xint()    {        int s = 1, c = xchar(), x = 0;        while (c <= 32) c = xchar();        if (c == '-') s = -1, c = xchar();        for (; '0' <= c && c <= '9'; c = xchar()) x = x * 10 + c - '0';        return x * s;    }    inline void xstring(char *s)    {        int c = xchar();        while (c <= 32) c = xchar();        for (; c > 32; c = xchar()) * s++ = c;        *s = 0;    }    inline void wchar(int x)    {        if (wpos == S) fwrite(wbuf, 1, S, stdout), wpos = 0;        wbuf[wpos ++] = x;    }    inline void wint(LL x)    {        if (x < 0) wchar('-'), x = -x;        char s[24];        int n = 0;        while (x || !n) s[n ++] = '0' + x % 10, x /= 10;        while (n--) wchar(s[n]);    }    inline void wstring(const char *s)    {        while (*s) wchar(*s++);    }    ~FastIO()    {        if (wpos) fwrite(wbuf, 1, wpos, stdout), wpos = 0;    }} io;typedef complex <double> Complex;void rader(Complex *y, int len) {    for(int i = 1, j = len / 2; i < len - 1; i++) {        if(i < j) swap(y[i], y[j]);        int k = len / 2;        while(j >= k) {j -= k; k /= 2;}        if(j < k) j += k;    }}void fft(Complex *y, int len, int op) {    rader(y, len);    for(int h = 2; h <= len; h <<= 1) {        double ang = op * 2 * PI / h;        Complex wn(cos(ang), sin(ang));        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(op == -1) for(int i = 0; i < len; i++) y[i] /= len;}int n, m, x;Complex x1[maxn], x2[maxn];int main(){    //scanf("%d%d", &n, &m);    n = io.xint();    m = io.xint();//    REP1(i, 0, n) a[i] = io.xint();//    REP1(i, 0, m) b[i] = io.xint();    int len = 1;    while(len < n * 2 || len < m * 2) len <<= 1;    REP1(i, 0, n){        x = io.xint();        x1[i] = Complex(x, 0);    }    REP1(i, n, len) x1[i] = Complex(0, 0);    REP1(i, 0, m){        x = io.xint();        x2[i] = Complex(x, 0);    }    REP1(i, m, len) x2[i] = Complex(0, 0);    fft(x1, len, 1);    fft(x2, len, 1);    REP1(i, 0, len) x1[i] = x1[i] * x2[i];    fft(x1, len, -1);    len = n + m - 1;    for(int i = 0; i < len - 1; i++) printf("%d ", (int) (x1[i].real() + 0.5));    printf("%d\n", (int) (x1[len - 1].real() + 0.5));    return 0;}