0_0_29468998_9255.cpp:1:10: error: #include expects "FILENAME" or <FILENAME>
#include #include #include #include using namespace std;typedef __int64 ll;const double pi = acos(-1.0);const int N = 10000+5;const double eps = 1e-6;const int MOD = 119; struct Complex { double x,y; Complex(double _x=0,double _y=0) :x(_x),y(_y) {} Complex operator + (const Complex& tt)const { return Complex(x+tt.x,y+tt.y); } Complex operator - (const Complex& tt)const { return Complex(x-tt.x,y-tt.y); } Complex operator * (const Complex& tt)const { return Complex(x*tt.x-y*tt.y,x*tt.y+y*tt.x); }};void FFT(Complex *a, int n, int rev) { // n是大于等于相乘的两个数组长度的2的幂次 // 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= MOD) x -= MOD;} void mul(int *tmp1, int *tmp2, int maxn) { //for(int i = 0;i < q; i++) printf("%d ", tmp1[i]); puts(""); //for(int i = 0;i < q; i++) printf("%d ", tmp2[i]); puts(""); for(int i = 0;i < maxn; i++) { a1[i].x = a1[i].y = a2[i].x = a2[i].y = 0; if(i < q) { a1[i].x = 1.0*tmp1[i]; a2[i].x = 1.0*tmp2[i]; } } FFT(a1, maxn, 1); FFT(a2, maxn, 1); for(int i = 0;i < maxn; i++) a1[i] = a1[i]*a2[i]; FFT(a1, maxn, -1); for(int i = 0;i < maxn; i++) Fuck[i] = (ll)(a1[i].x+eps)%MOD; for(int i = q*2-2;i >= q; i--) { if(Fuck[i]) { Fuck[i-q] = (Fuck[i-q] + Fuck[i]*b)%MOD; Fuck[i-p] = (Fuck[i-p] + Fuck[i]*a)%MOD; } } //printf("q = %d\n", q); for(int i = 0;i < q; i++) { tmp1[i] = (int)Fuck[i]; }} int ans[N], tmp[N], f[N]; int solve() { a %= MOD; b %= MOD; f[0] = 1; for(int i = 1;i < q; i++) { if(i-p < 0) f[i] = a+b; else if(i < q) f[i] = (a*f[i-p]+b)%MOD; } if(n < q) { return f[n]; } int maxn = 1; while(maxn <= (q-1)*2) maxn <<= 1; for(int i = 0;i < q; i++) ans[i] = tmp[i] = 0; ans[0] = 1; tmp[1] = 1; int fuck = n; while(n) { if(n&1) { mul(ans, tmp, maxn); } mul(tmp, tmp, maxn); n >>= 1; } int ret = 0; for(int i = 0;i < q; i++) { Add(ret, ans[i]*f[i]%MOD); } n = fuck; return ret;} int main() { while(input()) { printf("%d\n", solve()); } return 0;}
^
|
