A. Distance
B. Special Permutation
C. Chat Ban
D.X-Magic Pair
E. Messages
F：没看F，好难的样子
G. Max Sum Array

#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
#include <cmath>
#include <stack>
#include <map>
#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 = 1e5, M = 5e2;int main() 
{int t;scanf("%d", &t);while(t --){int a, b;scanf("%d%d", &a, &b);if(abs(a)+abs(b)&1)puts("-1 -1");else{int x = abs(a)>>1, y = (abs(a)+abs(b)>>1)-x;if(a<0)x=-x;if(b<0)y=-y;cout<<x<<' '<<y<<endl;}}return 0;
}

硬模拟

#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
#include <cmath>
#include <stack>
#include <map>
#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 = 1e5, M = 5e2;int main() 
{int t;scanf("%d", &t);while(t --){int n, a, b, l = 1, r = 2, pos[N]={0}, nl = 0, nr = 0;bool f = 0;scanf("%d%d%d", &n, &a, &b);for(int i = n;i > b;i --)pos[i] = l, nl++;if(!pos[a])pos[a] = l, nl++;for(int i = 1;i < a;i ++)f |= pos[i] == l, pos[i] = r, nr++;if(!pos[b]) f |= pos[b] == l, pos[b] = r, nr++;for(int i = a+1;i < b;i ++)if(!pos[i]){if(nl < n/2)pos[i] = l, nl++;else pos[i] = r;}if(f || nl!=n/2)puts("-1");else{for(int i = 1;i <= n;i ++)if(pos[i] == l)cout<<i<<' ';for(int i = 1;i <= n;i ++)if(pos[i] == r)cout<<i<<' ';puts("");}}return 0;
}

二分直接算

#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
#include <cmath>
#include <stack>
#include <map>
#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 = 1e5, M = 5e2;LL k, x;bool ch(LL l)
{LL sum = 0;if(l>k) sum = (1+k)*k/2 + (k+2*k-l-1)*(l-k)/2;else sum = (1+l)*l/2;return sum>=x;
}
int main() 
{int t;scanf("%d", &t);while(t --){scanf("%lld%lld", &k, &x);LL l = 1, r = 2*k-1;while(l < r){if(ch(mid)) r=mid;else l=mid+1;}cout<<l<<endl;}return 0;
}

队友写的，没搞懂

#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
#include <cmath>
#include <stack>
#include <map>
#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 = 1e5, M = 5e2;int main() 
{int t;scanf("%d", &t);while(t --){LL a, b, x;bool f = 0;scanf("%lld%lld%lld", &a, &b, &x);if(x<=max(a, b)){while(a>=x||b>=x&&a&&b){if(a<b)swap(a, b);if(a==x||b==x||(a-x)%b==0){f = 1;break;}a %= b;}}puts(f?"YES":"NO");}return 0;
}

 概率公式计算，假设有n张牌，其中里面有他要的，它可以抽k张牌，如果k>n那么这个人的期望就是n/n，否则是1 - $\left(\genfrac{}{}{0ex}{}{n-1}{k}\right)$/$\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ = k / n。
 然后可以枚举小于20的n枚举的时候k大于n的贡献要变成1。
 对于大于20的n来说答案就可以按照期望贡献对数字排序，再进行递推枚举计算。
具体看代码

#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
#include <cmath>
#include <stack>
#include <map>
#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, M = 5e2;int a[N][21]; 
PII b[N];
vector<int>ans;int main() 
{int n;scanf("%d", &n);for(int i = 1;i <= n;i ++){int x, y;scanf("%d%d", &x, &y);a[x][y]++;  //有个 数字为x，k等于y的人 }LL ma = 0, mn = 1;   // ma 是最大值， mn是最大值个数其实在ans-vector里有体现； for(int i = 1;i <= 20;i ++){      // 枚举n  for(int j = 1;j <= 2e5;j ++){ // 遍历每个树 int sum = 0;  // 当n等于i时数字i对于期望的贡献 for(int k = 1;k <= i;k ++)sum += a[j][k]*k;  // k ＜ n时 k = k， k ≥n时 k = n；  for(int k = i+1;k <= 20;k ++)sum += a[j][k]*i;b[j] = {-sum, j};   // 第一关键字取反按照从小往大排序的话第一个取反就是最大值； }sort(b+1, b+(int)2e5+1);  LL sum = 0;for(int x = 1;x <= i;x ++)sum += -b[x].first;if(sum*mn>ma*i){   // 分式化乘法 ans.clear();for(int x = 1;x <= i;x ++)ans.push_back(b[x].second);ma = sum; mn = i;}	}LL h = 0;for(int i = 1;i <= 20&&i <= n;i ++)h += -b[i].first;  // 这时候是先求出前二十个的和； for(int i = 21;i <= n;i ++) // 枚举21 - n 的个数取值； {h += -b[i].first;if(h*mn>ma*i)ma = h, mn = i;}if(mn > 20)for(int i = 1;i <= mn;i ++)ans.push_back(b[i].second);cout<<ans.size()<<endl;for(auto x: ans)cout<<x<<' ';cout<<endl;return 0;
}

  没人补G吗，假设有n个相等数字且位置为P1 P2 P3 …Pn，那么你计算之后就可以得到 ${\sum }_{i=1}^n$ ( 2 $\ast$ i - n -1) $\ast$ Pi

  那么再对于每个n都确定前面的 ( 2 $\ast$ i - n -1)，那么这个东西就由 Pi 确定，然后可以看出( 2 $\ast$ i - n -1)要么都是奇数要么都是偶数那么再把不连续的变成连续的(这步在代码里面有注释), 再挨个计算，数量就是( 2 $\ast$ i - n -1)相等的数量的阶乘的乘积

#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
#include <cmath>
#include <stack>
#include <set>
#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 = 2e6+10, mod = 1e9 + 7;
void mull(int &a, LL b){a = a*b%mod;return ;}
void add(int &a, LL b){a = (a+b)%mod;return ;}LL a[N], b[N], in[N];
// a ：-2 0 2 4 6 -> -1 0 1 2 3       n 是奇数 
// b ：-3 -1 1 3  -> -2  0 2 4 < a    n 是偶数 
LL se(LL a, int b){a%=mod;return b*(a+a+b-1)%mod*500000004%mod;}int main() 
{int n;scanf("%d", &n);in[0] = 1;for(int i = 1;i <= n;i ++){int x;scanf("%d", &x);LL *t = a, f = 0;if(x%2 == 0)t = b, f = 1;t[(1-x+f)/2+(int)1e6]++;t[(x-1+f)/2+(int)1e6+1]--;in[i] = i*in[i-1]%mod;}int ans = 0, sum = 1;LL head = 1;for(int i = 0;i <= 2e6;i ++){a[i] += a[i-1]; b[i] += b[i-1];add(ans, ((i-(int)1e6)*2-1)*se(head, b[i])%mod);head += b[i];add(ans, ((i-(int)1e6)*2)*se(head,   a[i])%mod);head += a[i];mull(sum, in[b[i]]*in[a[i]]%mod);}add(ans, mod);cout<<ans<<' '<<sum<<endl;return 0;
}