2021CCPC 威海 （dp + two points or NTT）
题解
\(E_{i}\)表示i次重开的期望值。
\(E_i = E_{i - 1} * P(random\_sum < E_{i - 1}) + E(random\_sum \geq E_{i - 1})\)
即如果当前随机的两张牌的和小于\(E_{i - 1}\)，则选择重开；否则不重开。有i次重开机会的期望就是重开的概率乘以i-1次重开的期望值，再加上不重开的期望。
开始想到一种NTT的做法，即用NTT算出两两之和的所有情况，因为\(max\{a_i\} \leq 10^6\)，所以两两的和不会超过\(2\times 10^6\)，时间复杂度为\(O(max\{a_i\}\log max\{a_i\} + max\{a_i\})\)。
但是这道题时限只给了1秒，我写的NTT最终还是T了。不过看到有同学写的FFT只用了600多ms就跑出来了，估计是NTT取模太慢了吧。
可以注意到这题k只给到了100，如果每次计算\(E_i\)时采用双指针的方式对重开的情况重新计算，时间复杂度为\(O(nk)\)。
但是NTT的做法也不是一无是处的，如果把这题的k改为\(10^5\)甚至更大，把时限稍微开大点，就只能NTT了。
code(NTT)
/*************************************************************************
> File Name: 1.cpp
> Author: Knowledge_llz
> Mail: 925538513@qq.com
> Blog: https://www.cnblogs.com/Knowledge-Pig/
************************************************************************/
#include<bits/stdc++.h>
#define LL long long
#define endl '\n'
using namespace std;
const int N = 2097152, mod = 998244353;
const double eps = 1e-6;
LL n, Q, k, a[N + 100], b[N + 100], c[N + 100], dp[N + 100], lim;
double E[N + 100];
LL qpow(LL x, LL y){
LL res = 1;
while(y){
if(y & 1) res = res * x % mod;
x = x * x % mod;
y >>= 1;
}
return res;
}
void ntt(LL *x, bool idft){
for(int i = 1; i < lim; ++i)
if(i < dp[i]) swap(x[i], x[dp[i]]);
for(int i = 2; i <= lim; i <<= 1){
LL w = qpow(3, (mod - 1) / i);
int len = i / 2;
if(idft) w = qpow(w, mod - 2);
for(int k = 0; k < lim; k += i){
LL p = 1;
for(int l = k; l < k + len; ++l){
LL tmp = p * x[len + l] % mod;
x[len + l] = (x[l] - tmp + mod) % mod;
x[l] = (x[l] + tmp) % mod;
p = p * w % mod;
}
}
}
}
int main(){
ios::sync_with_stdio(false); cin.tie(0);
#ifndef ONLINE_JUDGE
freopen("input.in", "r", stdin);
freopen("output.out","w", stdout);
#endif
cin >> n >> k >> Q;
for(int i = 1; i <= n; ++i){
cin >> a[i];
++b[a[i]];
}
for(lim = 1; lim < 2000000; lim <<= 1);
for(int i = 1; i < lim; ++i)
dp[i] = (dp[i >> 1] >> 1) | ((i & 1) ? lim >> 1 : 0);
ntt(b, 0);
for(int i = 0; i < lim; ++i) c[i] = b[i] * b[i] % mod;
ntt(c, 1);
LL inv = qpow(lim, mod - 2);
for(int i = 0; i < lim; ++i) c[i] = c[i] * inv % mod;
for(int i = 1; i <= n; ++i) --c[a[i] * 2];
LL sum = (n - 1ll) * n; double ans = 0;
for(int i = 0; i < lim; ++i) ans += c[i] * i;
E[0] = ans / sum; LL num = 0;
for(int i = 1, j = 0; i <= 100; ++i){
while((E[i - 1] - j) > eps){
ans += c[j] * (E[i - 1] - j);
num += c[j++];
}
E[i] = ans / sum;
ans += (E[i] - E[i - 1]) * num;
}
cout << fixed << setprecision(10) << E[k] << endl;
while(Q--){
int x, y, c;
cin >> x >> y >> c;
sum = a[x] + a[y];
if(!c) cout << "accept" << endl;
else if(fabs(sum - E[c - 1]) <= eps) cout << "both" << endl;
else if(sum - E[c - 1] > eps) cout << "accept" << endl;
else cout << "reselect" << endl;
}
return 0;
}
code(two points)
/*************************************************************************
> File Name: 1.cpp
> Author: Knowledge_llz
> Mail: 925538513@qq.com
> Blog: https://www.cnblogs.com/Knowledge-Pig/
************************************************************************/
#include<bits/stdc++.h>
#define LL long long
#define endl '\n'
using namespace std;
const int N = 3e5, mod = 998244353;
const double eps = 1e-6;
LL n, Q, k, a[N + 100], b[N + 100], c[N + 100], dp[N + 100], sum[N];
double E[N + 100];
int main(){
ios::sync_with_stdio(false); cin.tie(0);
#ifndef ONLINE_JUDGE
freopen("input.in", "r", stdin);
freopen("output.out","w", stdout);
#endif
cin >> n >> k >> Q;
double ans = 0; LL fm = (n - 1ll) * n;
for(int i = 1; i <= n; ++i){
cin >> a[i];
b[i] = a[i];
ans += a[i] * (n - 1ll) * 2ll;
}
sort(b + 1, b + n + 1);
for(int i = 1; i <= n; ++i) sum[i] = sum[i - 1] + b[i];
E[0] = ans / fm;
for(int i = 1; i <= 100; ++i){
double x = ans;
for(int l = 1, r = n; l <= n; ++l){
while(r && b[l] + b[r] > E[i - 1]) --r;
x -= (b[l] * r + sum[r]);
if(l <= r) x += 2 * b[l];
x += E[i - 1] * (r - (l <= r));
// cout << l << " " << r << " " << x << endl;
}
E[i] = x / fm;
}
cout << fixed << setprecision(10) << E[k] << endl;
while(Q--){
int x, y, c;
cin >> x >> y >> c;
LL s = a[x] + a[y];
if(!c) cout << "accept" << endl;
else if(fabs(s - E[c - 1]) <= eps) cout << "both" << endl;
else if(s - E[c - 1] > eps) cout << "accept" << endl;
else cout << "reselect" << endl;
}
return 0;
}
