CodeCraft-20 (Div. 2)
E.
令dp[i][status]表示第i位，状态为status的时候，最优解。
对于每一个i的时候，遍历所有status转移，然后判断一下这个时候k能否继续插入。
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdlib>
#include<climits>
#include<stack>
#include<vector>
#include<queue>
#include<set>
#include<bitset>
#include<map>
//#include<regex>
#include<cstdio>
#pragma GCC optimize(2)
#define up(i,a,b) for(int i=a;i<b;i++)
#define dw(i,a,b) for(int i=a;i>b;i--)
#define upd(i,a,b) for(int i=a;i<=b;i++)
#define dwd(i,a,b) for(int i=a;i>=b;i--)
//#define local
typedef long long ll;
typedef unsigned long long ull;
const double esp = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int inf = 1e9;
using namespace std;
ll read()
{
char ch = getchar(); ll x = 0, f = 1;
while (ch<'0' || ch>'9') { if (ch == '-')f = -1; ch = getchar(); }
while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
return x * f;
}
typedef pair<int, int> pir;
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lrt root<<1
#define rrt root<<1|1
const int N = 1e5 + 10;
ll dp[N][160];
int n, p, k;
struct node {
ll ai;
ll s[10];
bool operator<(const node a)const {
return ai > a.ai;
}
}ps[N];
int main()
{
n = read(), p = read(), k = read();
upd(i, 1, n)ps[i].ai = read();
upd(i, 1, n)
{
up(j, 0, p)ps[i].s[j] = read();
}
memset(dp, -1, sizeof(dp));
sort(ps + 1, ps + n + 1);
int status = (1 << p) - 1;
dp[0][0] = 0;
upd(i, 1, n)
{
upd(j, 0, status)
{
int cnt = 0;
up(tp, 0, p)
{
if (j >> tp & 1)cnt++;
}
int num = i - 1 - cnt;
if (num < k&&dp[i-1][j]!=-1)
{
dp[i][j] = dp[i - 1][j] + ps[i].ai;
}
else dp[i][j] = dp[i - 1][j];
up(tp, 0, p)
{
if (j >> tp & 1)
{
if (dp[i - 1][j ^ (1 << tp)] != -1)
{
dp[i][j] = max(dp[i][j], dp[i - 1][j ^ (1 << tp)] + ps[i].s[tp]);
}
}
}
}
}
cout << dp[n][status] << endl;
return 0;
}
F.
首先我们计算\(p_i\)*\(p_j\)出现的概率（i<j)。出现\(p_i\)*\(p_j\)那么表示i和j中间没有数字。对于所有的集合来说,有\(C_{n-(j-i+1)}^{0}\)+\(C_{n-(j-i+1)}^{1}\)+\(C_{n-(j-i+1)}^{2}\)+......+\(C_{n-(j-i+1)}^{n-(j-i+1)}\)，即\(2^{n-(j-i+1)}\)，这就是\(p_i\)\(p_j\)出现的次数。期望就是\(2^{-(j-i+1)}\)。那么对于整体而言，期望就是
\(\sum_{i=1}^{n}\)\(\sum_{j=i}^{n}\)\(p_i\)*\(p_j\)*\(2^{-(j-i+1)}\)
分组求和变成\(\sum_{i=1}^{n}\)\(p_i\)*\(2^{(i-1)}\)*\(\sum_{j=i}^{n}\)\(p_j\)*\(2^{-j}\)
利用线段树即可。分别保存tot=\(\sum_{i=1}^{n}\)\(\sum_{j=i}^{n}\)\(p_i\)*\(p_j\)*\(2^{-(j-i+1)}\),lcr=\(\sum_{i=1}^{n}\)\(p_i\)*\(2^{(i-1)}\),rc=\(\sum_{j=i}^{n}\)\(p_j\)*\(2^{-j}\)
和当前区间的k次方。注意合并的时候右区间需要关注左区间的次方。
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdlib>
#include<climits>
#include<stack>
#include<vector>
#include<queue>
#include<set>
#include<bitset>
#include<map>
//#include<regex>
#include<cstdio>
#pragma GCC optimize(2)
#define up(i,a,b) for(int i=a;i<b;i++)
#define dw(i,a,b) for(int i=a;i>b;i--)
#define upd(i,a,b) for(int i=a;i<=b;i++)
#define dwd(i,a,b) for(int i=a;i>=b;i--)
//#define local
typedef long long ll;
typedef unsigned long long ull;
const double esp = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int inf = 1e9;
using namespace std;
ll read()
{
char ch = getchar(); ll x = 0, f = 1;
while (ch<'0' || ch>'9') { if (ch == '-')f = -1; ch = getchar(); }
while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
return x * f;
}
typedef pair<int, int> pir;
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lrt root<<1
#define rrt root<<1|1
const int mod = 1e9 + 7;
const int N = 3e5 + 10;
ll quick_pow(ll k, ll m)
{
ll res = 1;
while (m)
{
if (m & 1)res = res * k%mod;
k = k * k%mod;
m >>= 1;
}
return res;
}
ll bi_pow[2 * N];
ll bi_inv_pow[2 * N];
struct p {
ll pi, id, pos;
bool operator<(const p a)const
{
return pi < a.pi;
}
}pw[2 * N];
int f_pos[2 * N];
vector<ll>vec;
int n, q;
void init()
{
bi_pow[0] = 1;
upd(i, 1, n + q)bi_pow[i] = bi_pow[i - 1] * 2ll % mod;
upd(i, 0, n + q)bi_inv_pow[i] = quick_pow(bi_pow[i], mod - 2);
}
struct node {
ll tot_val;
ll lc_val, rc_val;
ll nums;
node operator+(const node temp)const
{
node res;
node now = *(this);
if (temp.nums == 0)return now;
if (now.nums == 0)return temp;
res.tot_val = (now.tot_val + temp.tot_val + (now.lc_val*temp.rc_val%mod)*bi_inv_pow[now.nums] % mod) % mod;
res.lc_val = (now.lc_val + temp.lc_val * bi_pow[now.nums] % mod) % mod;
res.rc_val = (now.rc_val + temp.rc_val*(bi_inv_pow[now.nums])%mod) % mod;
res.nums = now.nums + temp.nums;
return res;
}
}tr[N<<3];
void pushup(int root)
{
tr[root] = tr[lrt] + tr[rrt];
}
void build(int l, int r, int root)
{
if (l == r)
{
if(pw[l].id<=n)
{
tr[root].tot_val = 0;
tr[root].lc_val = pw[l].pi;
tr[root].rc_val = pw[l].pi*bi_inv_pow[1] % mod;
tr[root].nums = 1;
}
else
{
tr[root].tot_val = tr[root].lc_val = tr[root].rc_val = tr[root].nums = 0;
}
return;
}
int mid = (l + r) >> 1;
build(lson);
build(rson);
pushup(root);
}
void update(int l, int r, int root, int pos,int val)
{
if (l == r)
{
if (val)
{
tr[root].tot_val = 0;
tr[root].lc_val = pw[l].pi;
tr[root].rc_val = pw[l].pi*bi_inv_pow[1] % mod;
tr[root].nums = 1;
}
else
{
tr[root].tot_val = tr[root].lc_val = tr[root].rc_val = tr[root].nums = 0;
}
return;
}
int mid = (l + r) >> 1;
if (pos <= mid)update(lson, pos, val);
else update(rson, pos, val);
pushup(root);
}
int pre[N * 2];
int main()
{
n = read();
upd(i, 1, n)
{
pw[i].pi = read(); pw[i].id = i; vec.push_back(pw[i].pi);
//mp[pw[i].pi]++;
}
q = read();
vector<p>query;
upd(i, 1, q) {
pw[n + i].pos = read();
pw[n + i].pi = read();
pw[n + i].id = i + n;
query.push_back(pw[n + i]);
//mp[pw[i + n].pi]++;
}
//int cnt = 0;
//for (auto k : mp)
//{
// cnt += k.second;
// mp2[k.first] = cnt;
//}
sort(vec.begin(), vec.end());
sort(pw + 1, pw + 1 + n + q);
init();
//upd(i, 1, n + q)cout << pw[i].id << endl;
upd(i, 1, n + q)f_pos[pw[i].id] = i;
build(1, n + q, 1);
int sum = n + q;
printf("%lld\n", tr[1].tot_val%mod);
for (auto k : query)
{
int pos = f_pos[k.id];
int pos2;
if (pre[k.pos]==0)
pos2 = f_pos[k.pos];
else pos2 = pre[k.pos];
pre[k.pos] = pos;
update(1, sum, 1, pos2, 0);
update(1, sum, 1, pos, 1);
printf("%lld\n", tr[1].tot_val%mod);
}
return 0;
}
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