A. Divan and a Store
B. Divan and a New Project
C. Divan and bitwise operations
D1. Divan and Kostomuksha (easy version)
D2. Divan and Kostomuksha (hard version)
E. Divan and a Cottage

 排序贪心

#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
#include <cmath>
#include <stack>
#include <queue>
#define mid (l+r>>1)
#define lowbit(x) (x&-x)
using namespace std;
typedef long long LL;
typedef pair<int,int> PII;
const int N = 2e5+10, mod = 1e9+7;int a[N];
int main() 
{int t;scanf("%d", &t);while(t --){int n, l, r, k, ans = 0;scanf("%d%d%d%d", &n, &l, &r, &k);for(int i = 1;i <= n;i ++)scanf("%d", a+i);sort(a+1, a+n+1);for(int i = 1;i <= n;i ++)if(a[i] >= l&&a[i] <= r && a[i] <=k)ans ++, k -= a[i];cout<<ans<<endl;}return 0;
}?

 排序贪心，起点放到原点，左右延伸不会超过1e6

#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
#include <cmath>
#include <stack>
#include <queue>
#define mid (l+r>>1)
#define lowbit(x) (x&-x)
using namespace std;
typedef long long LL;
typedef pair<int,int> PII;
const int N = 2e5+10, mod = 1e9+7;PII a[N];
int c[N];int main() 
{int t;scanf("%d", &t);while(t --){int n;scanf("%d", &n);for(int i = 1;i <= n;i ++)scanf("%d", &a[i].first), a[i].first = -a[i].first, a[i].second = i;sort(a+1, a+n+1);LL ans = 0;c[0] = 0;int p = 1;for(int i = 1;i <= n;i ++){int x = -a[i].first, y = a[i].second;c[y] = p;ans += (LL)x*2*abs(p);if(p > 0)p = -p;else p = -p+1; }cout<<ans<<endl;for(int i = 0;i <= n;i ++)cout<<c[i]<<' ';}return 0;
}

 假设二进制第i位有a个数是1， b个数是0，显然a+b = n， 这种情况下选出这位异或为一的情况数为 $2^b\ast (C_a^1+C_a^3+C_a^5+C_a^7......)$ ，其中后半部分由 $C_a^b=C_{a-1}^b+C_{a-1}^{b-1}$ 得 等于 $2^{a-1}$, 那么情况数就是 $2^{a+b-1}=2^{n-1}$ 是个常数，而题又说了肯定有符合条件的数列，那么就只看这位有没有，而不看有多少

考场上的错误做法，当成可以把那段都赋成1了

#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
#include <cmath>
#include <stack>
#include <queue>
#define mid (l+r>>1)
#define lowbit(x) (x&-x)
using namespace std;
typedef long long LL;
typedef pair<int,int> PII;
const int N = 2e5+10, mod = 1e9+7;PII a[N];
int in[N], c[32][N];int main() 
{int t;scanf("%d", &t);in[0] = 1; for(int i = 1;i <= 2e5;i ++)in[i] = in[i-1] * 2 %mod;while(t --){int n, m;scanf("%d%d", &n, &m);for(int i = 0;i < 31;i ++)for(int k = 0;k <= n;k ++)c[i][k] = 0;for(int i = 1;i <= m;i ++){int l, r, x;scanf("%d%d%d", &l, &r, &x);for(int j = 0;j < 31;j ++)if(x>>j&1)c[j][l]++, c[j][r+1]--;}for(int i = 0;i < 31;i ++)for(int k = 1;k <= n;k ++)c[i][k] += c[i][k-1];for(int i = 0;i < 31;i ++)for(int k = 1;k <= n;k ++){if(c[i][k]) c[i][k] = 1;c[i][k] += c[i][k-1];}LL ans = 0;for(int i = 0;i < 31;i ++){if(c[i][n])ans += (LL)(1<<i)*in[c[i][n]-1]%mod*in[n-c[i][n]]%mod, ans %= mod;}cout<<ans<<endl;}return 0;
}

正解 很漂亮

#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
#include <cmath>
#include <stack>
#include <queue>
#define mid (l+r>>1)
#define lowbit(x) (x&-x)
using namespace std;
typedef long long LL;
typedef pair<int,int> PII;
const int N = 2e5+10, mod = 1e9+7;
int in[N];int main() 
{int t;scanf("%d", &t);in[0] = 1; for(int i = 1;i <= 2e5;i ++)in[i] = in[i-1] * 2 %mod;while(t --){int n, m, ans = 0;scanf("%d%d", &n, &m);while(m --){int l, r, x;scanf("%d%d%d", &l, &r, &x);ans |= x;}printf("%d\n", (LL)ans*in[n-1]%mod);}return 0;
}

 $dp$ 求解 考虑状态转移方程 $dp[i]$表示 $i$ 的倍数放一起得到的最大价值，那么$dp[i]=max(dp[i],dp[j]+i\ast (num[i]-num[j])$, 其中 $j$ 是 $i$ 的倍数，$num[i]$ 是 $i$ 的倍数的个数

#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
#include <cmath>
#include <stack>
#include <queue>
#define mid (l+r>>1)
#define lowbit(x) (x&-x)
using namespace std;
typedef long long LL;
typedef pair<int,int> PII;
const int N = 5e6+10, mod = 1e9+7;int prim[N], idx, a[N], num[N];
LL ans[N];
bool dis[N];void ini()
{for(int i = 2;i <= 5e6;i ++){if(!dis[i])prim[idx++] = i;for(int j = 0;j < idx&&(LL)i*prim[j]<=5e6;j ++){dis[i*prim[j]] = 1; if(i%prim[j] == 0)break;}}return ;
}	int main() 
{int n;scanf("%d", &n);ini();for(int i = 1, x;i <= n;i ++)scanf("%d", &x), a[x] ++;for(int i = 1;i <= 5e6;i ++)for(int j = i;j <= 5e6;j += i)num[i] += a[j];for(int i = 5e6;i >= 1;i --)for(int j = 0;j < idx;j ++)if(i*prim[j] <= 5e6)ans[i] = max(ans[i], ans[i*prim[j]]+(LL)i*(num[i]-num[i*prim[j]]));else break;cout<<ans[1]<<endl;return 0;
}

 D2主要优化了求$num[i]$的过程, 还没搞懂，有人教教我吗，呜呜呜

 看了题解，温度一定是单调不递减的，就可以区间修改，线段树做吧。

#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
#include <cmath>
#include <stack>
#include <queue>
#define mid (l+r>>1)
#define lowbit(x) (x&-x)
using namespace std;
typedef long long LL;
typedef pair<int,int> PII;
const int N = 2e5+10, mod = 1e9+7;int ls[N<<6], rs[N<<6], lazy[N<<6], ma[N<<6], mi[N<<6], root, idx;void add_tage(int u, int f){ mi[u] += f; ma[u] += f; lazy[u] += f;return ;}
void push_up(int u){ mi[u] = min(mi[ls[u]], mi[rs[u]]); ma[u] = max(ma[ls[u]], ma[rs[u]]);return ;}
void new_node(int &u, int l, int r){if(!u) u = ++idx, mi[u] = l, ma[u] = r;}void push_down(int u, int l, int r)
{new_node(ls[u], l, mid), new_node(rs[u], mid+1, r);add_tage(ls[u], lazy[u]);add_tage(rs[u], lazy[u]);lazy[u] = 0;
}void modify(int &u, int l, int r, int T)
{new_node(u, l, r);if(mi[u] > T){ add_tage(u, -1); return ;}if(ma[u] < T){ add_tage(u, 1); return ;}if(mi[u] == T && ma[u] == T) return ;if(l == r) return ;push_down(u, l, r);modify(ls[u], l, mid, T);	modify(rs[u], mid+1, r, T);push_up(u);
}
int quary(int &u, int l, int r, int x)
{new_node(u, l, r);if(l == r)return mi[u];push_down(u, l, r);return x <= mid ? quary(ls[u], l, mid, x) : quary(rs[u], mid+1, r, x);
}int main() 
{int n, lastans = 0;scanf("%d", &n);for(int i = 1;i <= n;i ++){int t, k;scanf("%d%d", &t, &k);modify(root, 0, 1e9, t);while(k --){int x;scanf("%d", &x);x = (x + lastans)%((int)1e9+1);printf("%d\n", lastans = quary(root, 0, 1e9, x));	} }return 0;
}?