HDU 5829 Rikka with Subset

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。

虽然只是用FFT的原理，NTT的模板。

将代码中的A和B卷积，然后再乘上，各自对应的系数1/（2……k*(k-1）！）；

需要需用：费马小定理，费马素数 FFT等等。

代码：

#include<iostream>#include<algorithm>#include<cstdio>#include<vector>#include<cstring>#include<map>#include<queue>#include<stack>#include<string>#include<cstdlib>#include<ctime>#include<set>#include<math.h>using namespace std;typedef long long ll;const int nn = 400100;const int inf = (1 << 31);const ll mod = 998244353;const ll g = 3;//mod的原根，当且仅当 g^(MOD-1) = 1 % MOD#define pb push_back#define lson l,mid,rt<<1#define rson mid+1,r,rt<<1|1ll n,m,q;ll a[nn];ll inv[nn];//i的阶乘的倒数%modll fac[nn]; //i的阶乘ll f[nn];//记录答案ll A[nn];//2^(n-i)/i!;ll B[nn];//i!*a[i+1];ll inv2;ll ppow(ll x, ll n){ll ans = 1;while (n){if (n & 1) ans = ans*x%mod;x = x*x%mod;n >>= 1;}return ans;}void init(){inv2 = ppow(2ll, mod - 2);inv[1] = inv[0] = 1;fac[0] = fac[1] = 1;for (ll i = 2; i <= 1e5; i++){inv[i] = inv[i - 1] * ppow(i, mod - 2)%mod;fac[i] = fac[i - 1] * i%mod;}}void rader(ll y[], ll len){for (ll i = 1, j = len / 2; i < len-1; i++){if (i < j) swap(y[i], y[j]);ll k = len / 2;while (j >= k) {j -= k;k /= 2;}if (j < k)j += k;}}void ntt(ll y[], int len, int on) {rader(y, len);for (int h = 2; h <= len; h <<= 1) {ll wn = ppow(g, (mod - 1) / h); //****//if (on  == -1) wn = ppow(wn, mod - 2);for (int j = 0; j < len; j += h) {ll w = 1;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 = ppow(len, mod - 2);for (int i = 0; i < len; i++)y[i] = y[i] * t % mod;}}bool cmp(ll x, ll y){return x > y;}int main(){int t;scanf("%d", &t);init();while(t--){scanf("%lld", &n);memset(A, 0, sizeof(A));memset(B, 0, sizeof(B));memset(a, 0, sizeof(a));for (ll i = 1; i <= n; i++) scanf("%lld", &a[i]);sort(a + 1, a + n + 1,cmp);for (ll i = 0; i < n; i++) A[i] = inv[i] * ppow(2, n-i) % mod;for (ll i = 1; i <= n; i++) B[i] = fac[i - 1] * a[i] % mod;reverse(B + 1, B + n + 1);for (ll i = 0; i < n; i++) B[i] = B[i + 1];B[n] = 0;ll len = 1;while (len <= n * 2)len <<= 1;ntt(A, len, 1);ntt(B, len, 1);for (ll i = 0; i < len; i++) A[i] = A[i] * B[i] % mod;ntt(A, len, -1);ll r = inv2;for (ll i = 1; i <= n; i++){f[i] = r*inv[i - 1] % mod*A[n - i] % mod;f[i] = (f[i] + f[i - 1]) % mod;printf("%lld ", f[i]);r = r*inv2%mod;}puts("");}return 0;}