题目:http://acm.hdu.edu.cn/showproblem.php?pid=5730
DP式:\( f[i] = \sum\limits_{j=1}^{i} f[i-j] * a[j] \)
因为没有给 \( f[0] \) 赋初值,所以在递归底层令 \( f[l] += a[l] \)
注意多组数据清空数组;
读入 \( s[i] \) 时要取模!!
代码如下:
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> using namespace std; typedef long long ll; typedef double db; int const xn=(1<<18),mod=313; db const Pi=acos(-1.0); int n,rev[xn],f[xn],s[xn]; struct cpl{ db x,y; cpl(db xx=0,db yy=0):x(xx),y(yy) {} }a[xn],b[xn]; cpl operator + (cpl a,cpl b){return cpl(a.x+b.x,a.y+b.y);} cpl operator - (cpl a,cpl b){return cpl(a.x-b.x,a.y-b.y);} cpl operator * (cpl a,cpl b){return cpl(a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x);} int rd() { int ret=0,f=1; char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=0; ch=getchar();} while(ch>='0'&&ch<='9')ret=ret*10+ch-'0',ch=getchar(); return f?ret:-ret; } void fft(cpl *a,int tp,int lim) { for(int i=0;i<lim;i++) if(i<rev[i])swap(a[i],a[rev[i]]); for(int mid=1;mid<lim;mid<<=1) { cpl wn=cpl(cos(Pi/mid),tp*sin(Pi/mid)); for(int j=0,len=(mid<<1);j<lim;j+=len) { cpl w=cpl(1,0); for(int k=0;k<mid;k++,w=w*wn) { cpl x=a[j+k],y=w*a[j+mid+k]; a[j+k]=x+y; a[j+mid+k]=x-y; } } } if(tp==1)return; for(int i=0;i<lim;i++)a[i].x=a[i].x/lim; } int upt(int x){while(x>=mod)x-=mod; while(x<0)x+=mod; return x;} void work(int l,int r) { if(l==r){f[l]=upt(f[l]+s[l]); return;}//f[0]=0... int len=r-l+1,mid=((l+r)>>1); work(l,mid); int lim=1,L=0; while(lim<len)lim<<=1,L++; for(int i=0;i<lim;i++)rev[i]=((rev[i>>1]>>1)|((i&1)<<(L-1))); for(int i=l;i<=mid;i++)a[i-l].x=f[i],a[i-l].y=0;//y for(int i=mid-l+1;i<lim;i++)a[i].x=0,a[i].y=0; for(int i=0;i<lim;i++)b[i].x=s[i],b[i].y=0; fft(a,1,lim); fft(b,1,lim); for(int i=0;i<lim;i++)a[i]=a[i]*b[i]; fft(a,-1,lim); for(int i=mid+1;i<=r;i++)f[i]+=(ll)(a[i-l].x+0.5)%mod; work(mid+1,r); } int main() { while(1) { n=rd(); if(!n)return 0; memset(f,0,sizeof f); memset(s,0,sizeof s); for(int i=1;i<=n;i++)s[i]=rd()%mod;//%mod!! work(1,n); printf("%d\n",f[n]); } }