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两圆相交特征形以后会补上文字

高联难度几何题第一道BECF构成调和四边形GB、GE、GF、GC构成调和线束I、B、E、H构成调和点列∠EFG为直角FE为BFH的内角平分线//接下来我们从求证出发，寻求本题的欧式解法欲证∠BFE=∠HFE⇔∠DBH=∠DFH⇔DHFB四点共圆⇔∠DFB=∠DHB=∠DBH⇔DB=DH=DC⇔D是BCH外接圆的圆心⇔∠BDC=2∠BHC现在看来我们已经完成了本题证明的关键部分，剩下的就是倒角了：∠BHC=Π/2-∠BGC∠BDC=Π-2∠DBC=Π-2∠BGC得证.

起因//从大佬那抄的#include<cstdio>#include<algorithm>#include<vector>using std::vector;using std::swap;typedef long long ll;const int N=1e6+10;const ll P=998244353;const ll G=3;ll read(){ ll a=0;int op=1;char ch=getchar(); whil

压位高精在高精度运算中加入了压位思想。详情请看代码（以高精加法为例）#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#define p 8 //要压的数#define carry 100000000 //用于进位 using namespace std;const int N=100010;char s1[N],s2[N];int a[

#include<bits/stdc++.h>using namespace std;#define re register#define il inlineil int read(){ re int x=0,f=1;char c=getchar(); while(c<'0'||c>'9'){if(c=='-') f=-1;c=getchar();} while(c>='0'&&c<='9') x=(x<<3)

1.while (scanf("%d",&input)!=EOF){2.while((cin>>m)!=0)

luoguP5461赦免战俘#include<bits/stdc++.h>using namespace std;int n;int main(){ scanf("%d",&n); for(int i=0;i<(1<<n);i++){ for(int j=0;j<(1<<n);j++){ printf("%d ",(i|j)!=((1<<n)-1)?0:1);} printf("\n");} return 0;}

void write(int x){ if (x<0){ putchar('-'); x=-x; } if (x>9) write(x/10); putchar(x%10+'0');}int read(){ int x=0,f=1; char c=getchar(); while (c<'0'||c>'9'){ if (c=='-') f=-1; c=getchar(); } while (c>='0'&&c<='.

//扩展欧几里得求逆元 O(log n)void Exgcd(ll a, ll b, ll &x, ll &y) { if (!b) x = 1, y = 0; else Exgcd(b, a % b, y, x), y -= a / b * x;}//快速幂求逆元 O (log n)ll fpm(ll x, ll power, ll mod) { x %= mod; ll ans = 1; for (; power; power >&g

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