代码大全

代码大全

常数优化

编译器卡常优化(CF上比较有用)

#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

读入优化:

inline int getint() {
	register char ch;
	while(!isdigit(ch=getchar()));
	register int x=ch^'0';
	while(isdigit(ch=getchar())) x=(((x<<2)+x)<<1)+(ch^'0');
	return x;
}

mmap读入优化:

#include<sys/mman.h>
#include<sys/stat.h>
class MMapInput {
	private:
		char *buf,*p;
		int size;
	public:
		MMapInput() {
			register int fd=fileno(stdin);
			struct stat sb;
			fstat(fd,&sb);
			size=sb.st_size;
			buf=reinterpret_cast<char*>(mmap(0,size,PROT_READ,MAP_PRIVATE,fileno(stdin),0));
			p=buf;
		}
		char getchar() {
			return (p==buf+size||*p==EOF)?EOF:*p++;
		}
};
MMapInput mmi;
inline int getint() {
	register char ch;
	while(!isdigit(ch=mmi.getchar()));
	register int x=ch^'0';
	while(isdigit(ch=mmi.getchar())) x=(((x<<2)+x)<<1)+(ch^'0');
	return x;
}

数论

扩欧求逆元:

void exgcd(const int &a,const int &b,int &x,int &y) {
	if(!b) {
		x=1,y=0;
		return;
	}
	exgcd(b,a%b,y,x);
	y-=a/b*x;
}
inline int inv(const int &x) {
	int ret,tmp;
	exgcd(x,mod,ret,tmp);
	return (ret%mod+mod)%mod;
}

多项式

FFT

inline void init_ru(const int &n) {
	for(register int i=0;i<n;i++) {
		ru[i]=(comp){cos(2*pi*i/n),sin(2*pi*i/n)};
		iru[i]=conj(ru[i]);
	}
}
inline void dft(comp f[],comp w[],const int &n) {
	for(register int i=0,j=0;i<n;i++) {
		if(i>j) std::swap(f[i],f[j]);
		for(register int l=n>>1;(j^=l)<l;l>>=1);
	}
	for(register int i=2;i<=n;i<<=1) {
		const int m=i>>1;
		for(register int j=0;j<n;j+=i) {
			for(register int k=0;k<m;k++) {
				const comp z=f[j+m+k]*w[n/i*k];
				f[j+m+k]=f[j+k]-z;
				f[j+k]+=z;
			}
		}
	}
}

FWT异或:

inline void fwt(int a[]) {
	for(register int j=0;j<len;j++) {
		for(register int i=0;i<n;i++) {
			if(i>>j&1) continue;
			const int x=a[i],y=a[(1<<j)|i];
			a[i]=(x+y)%mod;
			a[(1<<j)|i]=(x-y+mod)%mod;
		}
	}
}

字符串

后缀数组求sa[i]rank[i]

inline bool cmp(const int &i,const int &j) {
	if(rank[i]!=rank[j]) return rank[i]<rank[j];
	const int ri=i+k<=n?rank[i+k]:-1;
	const int rj=j+k<=n?rank[j+k]:-1;
	return ri<rj;
}
inline void suffix_sort() {
	for(register int i=0;i<=n;i++) {
		sa[i]=i;
		rank[i]=s[i];
	}
	for(k=1;k<=n;k<<=1) {
		std::sort(&sa[0],&sa[n]+1,cmp);
		tmp[sa[0]]=0;
		for(register int i=1;i<=n;i++) {
			tmp[sa[i]]=tmp[sa[i-1]]+!!cmp(sa[i-1],sa[i]);
		}
		std::copy(&tmp[0],&tmp[n]+1,rank);
	}
}

后缀数组求lcp[i]

inline void init_lcp() {
	for(register int i=0,h=0;i<n;i++) {
		if(h>0) h--;
		const int j=sa[rank[i]-1];
		while(j+h<n&&i+h<n&&s[j+h]==s[i+h]) h++;
		lcp[rank[i]-1]=h;
	}
}

计算几何:

自适应辛普森法:

inline double simpson(const double &a,const double &b) {
	const double c=(a+b)/2;
	return (F(a)+4*F(c)+F(b))*(b-a)/6;
}
inline double asr(const double &a,const double &b,const double &eps,const double &s) {
	const double c=(a+b)/2,ls=simpson(a,c),rs=simpson(c,b);
	if(fabs(ls+rs-s)<15*eps) return ls+rs+(ls+rs-s)/15;
	return asr(a,c,eps/2,ls)+asr(c,b,eps/2,rs);
}
posted @ 2018-06-08 09:58  skylee03  阅读(...)  评论(...编辑  收藏