HDU - 5829：Rikka with Subset （NTT）

As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them:

Yuta has n numbers A[1]~A[n] and a number K. For any none empty subset S of the numbers, the value of S is equal to the sum of the largest min(|S|,k) numbers in S. The value of the array A is equal to the sum of the value of all none empty subset of the numbers.

Now Yuta shows the n numbers, And he wants to know the value of the array for each K in [1,n].

It is too difficult for Rikka. Can you help her?

InputThe first line contains a number t(1<=t<=10), the number of the testcases.

For each testcase, the first line contains a number n(1<=n<=100000), the number of numbers Yuta has. The second line contains n number A[1]~A[n](0<=A[i]<=10^9).OutputFor each testcase, print a line contains exactly n numbers, the ith number is the value of the array when K=i. The answer may be very large, so you only need to print the answer module 998244353.

Sample Input

2
3
1 1 1
5
1 2 3 4 5

Sample Output

7 11 12
129 201 231 239 240

题意：给定一个数组，F(k)表示所有集合s的前min(K,s)大之和。求所有F(k)。

思路：先得到方程f(x)，然后一般来说一个组合数*一个指数，可以直接转化一下用NTT加速；或者用第二类斯特林转化，再套NTT或FFT卷积。

 

 

关键在于找到某两个系数之和为定值，然后分别以其为“基”构造函数，然后取卷积这两个函数。

#include<bits/stdc++.h>
#define rep(i,x,y) for(int i=x;i<=y;i++)
using namespace std;
#define ll long long
#define MOD Mod
const int G=3;
const int maxn=268576;
const int Mod=998244353;
int qpow(int v,int p)
{
    int ans=1;
    for(;p;p>>=1,v=1ll*v*v%Mod)
      if(p&1)ans=1ll*ans*v%Mod;
    return ans;
}
void rader(int y[], int len) {
    for(int i=1,j=len/2;i<len-1;i++) {
        if(i<j) swap(y[i],y[j]);
        int k=len/2;
        while(j>=k) j-=k,k/=2;
        if(j<k) j+=k;
    }
}
void NTT(int y[],int len,int opt) {
    rader(y,len);
    for(int h=2;h<=len;h<<=1) {
        int wn=qpow(G,(MOD-1)/h);
        if(opt==-1) wn=qpow(wn,Mod-2);
        for(int j=0;j<len;j+=h) {
            int w=1;
            for(int k=j;k<j+h/2;k++) {
                int u=y[k];
                int t=(ll)w*y[k+h/2]%MOD;
                y[k]=(u+t)%MOD;
                y[k+h/2]=(u-t+MOD)%MOD;
                w=(ll)w*wn%MOD;
            }
        }
    }
    if(opt==-1) {
        int t=qpow(len,MOD-2);
        for(int i=0;i<len;i++) y[i]=(ll)y[i]*t%MOD;
    }
}
int inv[maxn],A[maxn],B[maxn],a[maxn],f[maxn],p2[maxn];
int main() {
    int T,N;
    f[0]=inv[0]=p2[0]=1;
    rep(i,1,100000) p2[i]=(ll)p2[i-1]*2%Mod;
    rep(i,1,100000) f[i]=(ll)f[i-1]*i%Mod;
    inv[100000]=qpow(f[100000],Mod-2);
    for(int i=100000-1;i>=0;i--) inv[i]=(ll)inv[i+1]*(i+1)%Mod;
    scanf("%d",&T);
    while(T--){
        scanf("%d",&N);
        int len=1; while(len<=N*2) len<<=1;
        rep(i,0,len) A[i]=B[i]=0;
        rep(i,1,N) scanf("%d",&a[i]);
        sort(a+1,a+N+1); reverse(a+1,a+N+1);
        rep(i,0,N-1){
             A[i]=inv[i];
             B[i]=(ll)f[N-i-1]*p2[i]%Mod*a[N-i]%Mod;
        }

        NTT(A,len,1); NTT(B,len,1);
        rep(i,0,len-1) A[i]=(ll)A[i]*B[i]%Mod; //乘完，不能只乘到N
        NTT(A,len,-1); int ans=0;
        rep(i,1,N){
            (ans+=(ll)inv[i-1]*A[N-i]%Mod)%=Mod;
            printf("%d ",ans);
        }
        puts("");
    }
    return 0;
}

 

It is your time to fight！