HDU

JRY wants to drag racing along a long road. There are nn sections on the road, the ii-th section has a non-negative integer length sisi. JRY will choose some continuous sections to race (at an unbelievable speed), so there are totally n(n+1)2n(n+1)2 different ways for him to ride. If JRY rides across from the ii-th section to the jj-th section, he would gain j−i+1j−i+1 pleasure. Now JRY wants to know, if he tries all the ways whose length is ss, what's the total pleasure he can get. Please be aware that in the problem, the length of one section could be zero, which means that the length is so trivial that we can regard it as 00.

InputThe first line of the input is a single integer T (T=5)T (T=5), indicating the number of testcases. 

For each testcase, the first line contains one integer nn. The second line contains nnnon-negative integers, which mean the length of every section. If we denote the total length of all the sections as ss, we can guarantee that 0≤s≤500000≤s≤50000 and 1≤n≤1000001≤n≤100000. 
OutputFor each testcase, print s+1s+1 lines. The single number in the ii-th line indicates the total pleasure JRY can get if he races all the ways of length i−1i−1. 
Sample Input

2
3
1 2 3
4
0 1 2 3

Sample Output

0
1
1
3
0
2
3
1
3
1
6
0
2
7

题意：总区间中有n个数(n<=100000)，求每种区间和，累加出所对应的区间长度(j-i+1)和；

思路：可以通过母函数得到方程，然后FFT求。 也有一种暴力一点的方式，分治+FFT：即求跨过每个点的区间的贡献。 然后两次FFT分别算出左边和右边的贡献。

（4900ms卡过去了。要用long double。

母函数的方法可以看：https://blog.csdn.net/kyleyoung_ymj/article/details/51712329

#include<bits/stdc++.h>
#define ll long long
#define double long double
#define rep(i,x,y) for(int i=x;i<=y;i++)
#define rep2(i,x,y) for(int i=x;i>=y;i--)
using namespace std;
const int maxn=500010;
const double pi=acos(-1.0);
struct cp
{
    double r,i;
    cp(){}
    cp(double rr,double ii):r(rr),i(ii){}
    cp operator +(const cp&x)const{return cp(r+x.r,i+x.i);}
    cp operator -(const cp&x)const{return cp(r-x.r,i-x.i);}
    cp operator *(const cp&x)const{return cp(r*x.r-i*x.i,i*x.r+r*x.i);}
};
ll ans[maxn];int s[maxn],sum[maxn],R[maxn],n;
cp a[maxn],b[maxn],W,w,p;
void read(int &x){
    x=0; char c=getchar();
    while(c>'9'||c<'0') c=getchar();
    while(c>='0'&&c<='9') x=x*10+c-'0',c=getchar();
}
inline void fft(cp*c,int t)
{
    int i,j,k;
    for(i=0;i<n;i++) R[i]<i?swap(c[R[i]],c[i]),0:0;
    for(i=1;i<n;i<<=1)
     for(j=0,W={cos(pi/i),sin(pi/i)*t};j<n;j+=i<<1)
      for(k=0,w={1,0};k<i;k++,w=w*W)
       p=c[j+k+i]*w,c[j+k+i]=c[j+k]-p,c[j+k]=c[j+k]+p;
}
void solve(int l,int r)
{
    if(l==r){ ans[s[l]]++; return ;}
    int mid=(l+r)>>1 ,delta=sum[r]-sum[l-1];
    solve(l,mid); solve(mid+1,r);
    for(n=1;(n-1)<=delta;n<<=1);
    rep(i,1,n-1) R[i]=R[i>>1]>>1|(i&1?n>>1:0);

    rep2(i,mid,l) a[sum[mid]-sum[i-1]].r+=mid-i+1;
    rep(i,mid+1,r) ++b[sum[i]-sum[mid]].r;
    fft(a,1); fft(b,1);
    rep(i,0,n-1) a[i]=a[i]*b[i];
    fft(a,-1);
    rep(i,0,delta) ans[i]+=(ll)((a[i].r)/n+0.5);
    rep(i,0,n) a[i]=b[i]=cp(0.,0.);

    rep2(i,mid,l) ++a[sum[mid]-sum[i-1]].r;
    rep(i,mid+1,r) b[sum[i]-sum[mid]].r+=i-mid;
    fft(a,1); fft(b,1);
    rep(i,0,n-1) a[i]=a[i]*b[i];
    fft(a,-1);
    rep(i,0,delta) ans[i]+=(ll)((a[i].r)/n+0.5);
    rep(i,0,n) a[i]=b[i]=cp(0.,0.);
}
int main()
{
    int N,T;
    scanf("%d",&T);
    while(T--){
        read(N);
        rep(i,1,N) read(s[i]),sum[i]=sum[i-1]+s[i];
        rep(i,0,sum[N]) ans[i]=0;
        solve(1,N);
        rep(i,0,sum[N]) printf("%lld
",ans[i]);
    }
    return 0;
}