学习：https://blog.csdn.net/VictoryCzt/article/details/81005812
模板：https://blog.csdn.net/ransln/article/details/79462659
只是简单地学习了下，时间紧迫，主要还是找了个模板。
例题：
Rikka with Subset
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1343 Accepted Submission(s): 442
Problem Description
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?
Input
The 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).
Output
For 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： 给一个数组A[n], 要求出所有的T[k]，T[i]指A数组的所有子集中前k大的和的和。答案对998244353取模。
思路：
参考了https://blog.csdn.net/cqu_hyx/article/details/52194696
代码：
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int Max=5e5+100;
const ll Mod = 998244353; // 模数
const ll SpMul = 3; // 原根
ll qpow(ll a, ll k) {
ll res = 1LL;
while (k > 0) {
if (k & 1)res = res * a % Mod;
a = a * a % Mod;
k >>= 1;
}
return res;
}
void Change(ll 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(ll y[], int len, int on) {
Change(y, len);
for (int h = 2; h <= len; h <<= 1) {
ll wn = qpow(SpMul, (Mod - 1) / h);
if (on == -1)wn = qpow(wn, Mod - 2);
for (int j = 0; j < len; j += h) {
ll w = 1LL;
for (int k = j; k < j + h / 2; k++) {
ll u = y[k];
ll t = w * y[k + h / 2] % Mod;
y[k] = (u + t) % Mod;
y[k + h / 2] = (u - t + Mod) % Mod;
w = w * wn % Mod;
}
}
}
if (on == -1) {
ll t = qpow(len, Mod - 2);
for (int i = 0; i < len; i++)
y[i] = y[i] * t % Mod;
}
}
ll inv2;
ll pow2[Max];
ll invpow2[Max];
ll a[Max], b[Max];
ll A[Max];
ll inv[Max],fac[Max],invf[Max]; //inv[i]为i的逆元，fac[i]为阶乘i!，invf[i]为阶乘i!的逆元
void init()
{
inv[1]=1;
for (int i=2;i<Max;i++)
inv[i]=((Mod-Mod/i)*inv[Mod%i])%Mod;
fac[0] = 1;
for(int i = 1;i<Max;i++)
fac[i] = (fac[i-1] * (ll)i)%Mod;
invf[0] = 1;
for(int i = 1;i<Max;i++)
invf[i] = (invf[i-1]*inv[i])%Mod;
inv2 = qpow(2,Mod - 2);
pow2[0] = 1;
invpow2[0] = 1;
for(int i = 1;i<Max;i++)
{
pow2[i] = (pow2[i-1] * 2)%Mod;
invpow2[i] = (invpow2[i-1] * inv2)%Mod;
}
}
int main()
{
init();
int t;
scanf("%d",&t);
while(t--)
{
int n;
scanf("%d",&n);
int len = 1;
while(len<n*2)len<<=1;
for(int i = 1;i<=n;i++)
scanf("%lld",&A[i]);
sort(A+1,A+1+n,greater<ll>());
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
for(int i = 0;i<n;i++)
a[i] = (pow2[n-i] * invf[i])%Mod;
for(int i = 0;i<n;i++)
b[i] = (fac[n-i-1] * A[n-i])%Mod;
NTT(a,len,1);
NTT(b,len,1);
for(int i = 0;i<len;i++)
a[i] = (a[i] * b[i])%Mod;
NTT(a,len,-1);
ll ans = 0;
ll tmp;
for(int i = 1;i<=n;i++)
{
tmp = ((invpow2[i] * invf[i-1])%Mod * a[n-i])%Mod;
ans = (ans + tmp)%Mod;
printf("%lld ",ans);
}
puts("");
}
}