hdu 5830 FFT + cdq分治
Shell Necklace
Time Limit: 16000/8000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 647 Accepted Submission(s): 287
Problem Description
Perhaps the sea‘s definition of a shell is the pearl. However, in my view, a shell necklace with n beautiful shells contains the most sincere feeling for my best lover Arrietty, but even that is not enough.
Suppose the shell necklace is a sequence of shells (not a chain end to end). Considering i continuous shells in the shell necklace, I know that there exist different schemes to decorate the i shells together with one declaration of love.
I want to decorate all the shells with some declarations of love and decorate each shell just one time. As a problem, I want to know the total number of schemes.
Suppose the shell necklace is a sequence of shells (not a chain end to end). Considering i continuous shells in the shell necklace, I know that there exist different schemes to decorate the i shells together with one declaration of love.
I want to decorate all the shells with some declarations of love and decorate each shell just one time. As a problem, I want to know the total number of schemes.
Input
There are multiple test cases(no more than 20 cases and no more than 1 in extreme case), ended by 0.
For each test cases, the first line contains an integer n, meaning the number of shells in this shell necklace, where 1≤n≤105. Following line is a sequence with nnon-negative integer a1,a2,…,an, and ai≤107 meaning the number of schemes to decorate i continuous shells together with a declaration of love.
For each test cases, the first line contains an integer n, meaning the number of shells in this shell necklace, where 1≤n≤105. Following line is a sequence with nnon-negative integer a1,a2,…,an, and ai≤107 meaning the number of schemes to decorate i continuous shells together with a declaration of love.
Output
For each test case, print one line containing the total number of schemes module 313(Three hundred and thirteen implies the march 13th, a special and purposeful day).
Sample Input
3
1 3 7
4
2 2 2 2
0
Sample Output
14
54
/* hdu 5830 FFT + cdq分治 problem: 已知长度为i的shells有a[i]种. 求组成长度为n的方案数 f[i]=∑(f[i - j] * a[j]), j∈[1, i]; 让你求f[n] % Z 学习参考:http://blog.csdn.net/snowy_smile/article/details/52020971 solve: 首先卷积求出来之后坐标和相等的在同一列. a1 a2 a3 b1 b2 b3 -->> a1*b1 a2*b1 a3*b1 | a1*b2 a2*b2 | a3*b2 a1*b3 | a2*b3 a3*b3 所以可以解决多项式为f[i]=∑(f[i - j] * a[j])这种的问题.但是如果直接暴力的话 必需要n次fft递推出f[n]. 通过上面那个卷积公式可以发现当我们计算f[4]的时候,已经把后面一部分的答案计算了出来. 所以我们可以在计算[l,mid]的时候处理出f[l,mid]对[mid,r]的所有贡献. 那么剩下的就只需要在 [mid,r]中处理就行了.也就是CDQ分治了 hhh-2016-09-22 21:21:08 */ #pragma comment(linker,"/STACK:124000000,124000000") #include <algorithm> #include <iostream> #include <cstdlib> #include <cstdio> #include <cstring> #include <vector> #include <map> #include <math.h> #define lson i<<1 #define rson i<<1|1 #define ll long long #define clr(a,b) memset(a,b,sizeof(a)) #define key_val ch[ch[root][1]][0] using namespace std; const int maxn = 1 << 18; const int inf = 0x3f3f3f3f; const int mod = 313; const double eps = 1e-7; template<class T> void read(T&num) { char CH; bool F=false; for(CH=getchar(); CH<'0'||CH>'9'; F= CH=='-',CH=getchar()); for(num=0; CH>='0'&&CH<='9'; num=num*10+CH-'0',CH=getchar()); F && (num=-num); } int stk[70], tp; template<class T> inline void print(T p) { if(!p) { puts("0"); return; } while(p) stk[++ tp] = p%10, p/=10; while(tp) putchar(stk[tp--] + '0'); putchar('\n'); } const double PI = acos(-1.0); struct Complex { double x,y; Complex(double _x = 0.0,double _y = 0.0) { x = _x; y = _y; } Complex operator-(const Complex &b)const { return Complex(x-b.x,y-b.y); } Complex operator+(const Complex &b)const { return Complex(x+b.x,y+b.y); } Complex operator*(const Complex &b)const { return Complex(x*b.x-y*b.y,x*b.y+y*b.x); } }; 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]); k = len/2; while(j >= k) { j-=k; k/=2; } if(j < k) j+=k; } } 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].x /= len; } } double dis(int a,int b) { return sqrt(a*a + b*b); } Complex a[maxn]; Complex b[maxn]; int ans[maxn]; int ta[maxn]; void cal(int l,int r) { if(l == r) { ans[l] = (ans[l]+ta[l])%mod; return; } int mid = (l + r) >> 1; cal(l,mid); int len1 = mid-l + 1; int len2 = r-l + 1; int len = 1; while(len < (len1 + len2)) len <<= 1; for(int i = 0;i < len1;i++) a[i] = ans[l+i]; for(int i = len1;i < len;i++) a[i] = 0; fft(a,len,1); for(int i = 0;i < len2;i++) b[i] = ta[i]; for(int i = len2;i < len;i++) b[i] = 0; fft(b,len,1); for(int i = 0;i < len;i++) a[i] = a[i] * b[i]; fft(a,len,-1); for(int i = mid + 1;i <= r ;i++) ans[i] = (ans[i] + int(a[i-l].x + 0.5))%mod; cal(mid+1,r); } int main() { int n; while(scanf("%d",&n ) != EOF && n) { memset(ans,0,sizeof(ans)); for(int i = 1; i <= n;i++){ scanf("%d",&ta[i]); ta[i] %= mod; } cal(1,n); print(ans[n]); } return 0; }