CodeForces

题意：有N种物品，每种物品有价值(a_i),每种物品可选任意多个，求拿k件物品，可能损失的价值分别为多少。
分析：相当于求((a_1+a_2+...+a_n)^k)中，有哪些项的系数不为0.做k次FFT求卷积求卷积肯定爆炸，考虑用分治的形式计算，因为中间计算的时候会重复计算一些幂次，所以用记忆化搜索的形式，保留计算结果。
因为只要计算出哪些项不为0，所以卷积之后求结果时，系数非0项用1作系数即可，否则分分钟炸精度。
当然也可以用快速幂求解#.

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int MAXN = 2e6 + 10;
const double PI = acos(-1.0);
struct Complex{
    double x, y;
    inline Complex operator+(const Complex b) const {
        return (Complex){x +b.x,y + b.y};
    }
    inline Complex operator-(const Complex b) const {
        return (Complex){x -b.x,y - b.y};
    }
    inline Complex operator*(const Complex b) const {
        return (Complex){x *b.x -y * b.y,x * b.y + y * b.x};
    }
} va[MAXN * 2 + MAXN / 2], vb[MAXN * 2 + MAXN / 2];
int lenth = 1, rev[MAXN * 2 + MAXN / 2];
int N, M;   // f 和 g 的数量
    // f g和 的系数
    // 卷积结果
    // 大数乘积
int f[MAXN],g[MAXN];
vector<LL> conv;
vector<LL> multi;
//f g
void init()
{
    int tim = 0;
    lenth = 1;
    conv.clear(), multi.clear();
    memset(va, 0, sizeof va);
    memset(vb, 0, sizeof vb);
    while (lenth <= N + M - 2)
        lenth <<= 1, tim++;
    for (int i = 0; i < lenth; i++)
        rev[i] = (rev[i >> 1] >> 1) + ((i & 1) << (tim - 1));
}

void FFT(Complex *A, const int fla)
{
    for (int i = 0; i < lenth; i++){
        if (i < rev[i]){
            swap(A[i], A[rev[i]]);
        }
    }
    for (int i = 1; i < lenth; i <<= 1){
        const Complex w = (Complex){cos(PI / i), fla * sin(PI / i)};
        for (int j = 0; j < lenth; j += (i << 1)){
            Complex K = (Complex){1, 0};
            for (int k = 0; k < i; k++, K = K * w){
                const Complex x = A[j + k], y = K * A[j + k + i];
                A[j + k] = x + y;
                A[j + k + i] = x - y;
            }
        }
    }
}
void getConv(){             //求多项式
    init();
    for (int i = 0; i < N; i++)
        va[i].x = f[i];
    for (int i = 0; i < M; i++)
        vb[i].x = g[i];
    FFT(va, 1), FFT(vb, 1);
    for (int i = 0; i < lenth; i++)
        va[i] = va[i] * vb[i];
    FFT(va, -1);
    for (int i = 0; i <= N + M - 2; i++)
        conv.push_back((LL)(va[i].x / lenth + 0.5)>0?1:0);
}


int base[MAXN];
vector<LL> mi[1005];
int num;
bool make[1005];

vector<LL> dfs(int i,int j,int cc){
    if(i==j){
        vector<LL> res;
        for(int i=0;i<num;++i){
            res.push_back(base[i]);
        }
        make[cc] = true;
        mi[cc] = res;
        return res;
    }
    vector<LL> L,R;
    int m = (i+j)>>1;
    if(make[m-i+1]) L = mi[m-i+1];
    else L = dfs(i,m,m-i+1);
    if(make[j-m]) R = mi[j-m];
    else R = dfs(m+1,j,j-m);
    N = L.size();
    M = R.size();
    for(int i=0;i<N;++i) f[i] = L[i];
    for(int i=0;i<M;++i) g[i] = R[i];
    getConv();
    make[cc] = true;
    mi[cc] = conv;
    return conv;
}

int cnt[MAXN];

int main()
{
    int n,k;
    scanf("%d %d",&n, &k);
    int mx = -1;
    int mn = 100005;
    for(int i=1,tmp;i<=n;++i){
        scanf("%d",&tmp);
        cnt[tmp]++;
        mn = min(mn,tmp);
        mx = max(mx,tmp);
    }
    num = mx+1;
    for(int i=0;i<num;++i){
        base[i] = cnt[i];
    }
    conv = dfs(1,k,k);
    int sz = conv.size();
    for(int i=k*mn;i<sz;++i){
        if(!conv[i]) continue;
        printf("%d%c",i,i==sz-1?'
':' ');
    }
    return 0;
}