题目大意思:
求f[i] = sum(f[j] * a[i-j]) , a为题目给出。f[0]=1
然后后用CDQ分治,分治后用FFT。
要点:FFT的double精度有限……这题读入的时候要先mod再计算
第二:给kaungbin的FFT模板略加整理,更加好用一些。PS:多加一个整理函数,速度多300ms?!
#include<iostream> #include<cstdio> #include<cstdlib> #include<algorithm> #include<cstring> #include <cmath> using namespace std; const double PI = acos(-1.0); const int maxn = 100010 * 4; 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); } }; /* * 进行FFT和IFFT前的反转变换。 * 位置i和 (i二进制反转后位置)互换 * len必须去2的幂 */ 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]); //交换互为小标反转的元素,i<j保证交换一次 //i做正常的+1,j左反转类型的+1,始终保持i和j是反转的 k = len/2; while( j >= k) { j -= k; k /= 2; } if(j < k) j += k; } } /* * 做FFT * len必须为2^k形式, * on==1时是DFT,on==-1时是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; } complex x1[maxn], x2[maxn]; void getfft(int arg1[], int len1, int arg2[], int len2, int fft_out[])//模板:第一个数组,数组元素数量。 第二个数组,数组元素数量,输出数组 { int len = 1; while(len < len1*2 || len < len2*2)len<<=1; for(int i = 0;i < len1;i++) x1[i] = complex(arg1[i] , 0); for(int i = len1;i < len;i++) x1[i] = complex(0 , 0); for(int i = 0;i < len2;i++) x2[i] = complex(arg2[i] , 0); for(int i = len2;i < len;i++) x2[i] = complex(0, 0); fft(x1,len,1); fft(x2,len,1); for(int i = 0;i < len;i++) x1[i] = x1[i]*x2[i]; fft(x1,len,-1); for(int i = 0;i < len;i++) fft_out[i] = (int)(x1[i].r+0.5); } int n; const int mod = 313; int a[maxn], f[maxn], sum[maxn]; void init() { memset(f, 0, sizeof(f)); for (int i = 1; i <= n; ++i) { scanf("%lld", &a[i]); a[i] %= 313; } } void solve(int l ,int r) { if (l == r) { f[l] = (f[l] + a[l]) % mod; //f[l]一定已经求出了 return; } int mid = l + (r-l)/2; solve(l, mid); int len1 = mid -l + 1; int len2 = r - l; getfft(&f[l], len1, &a[1], len2, sum); for (int i = mid + 1; i <= r; ++ i) f[i] = (f[i] + sum[i-mid-1 + len1 -1]) % mod; solve(mid + 1, r); } void doit() { solve(1, n); printf("%lld\n", f[n]); } int main() { while (~scanf("%d", &n)) { if (!n) break; init(); doit(); } return 0; }