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写给自己看的模板

这是写给自己看的模板……当然有需要的人也可以直接拿走

1、封装好的高精度模板

注意:算法的主要耗时在于高精度乘法、除法和乘方。其中高精度乘方有快速幂优化,常数大致是乘法的5~10倍。除法是用二分写的效率较低,大概是乘法的20~30倍,能不用尽量不用

 

#define mx 300struct gaojing{     int len;     int a[mx+10];}zero,one;inline void set0(gaojing &s)//高精清零{    s.len=1;    for (int i=1;i<=mx+5;i++)s.a[i]=0;}inline void inputn(gaojing &a)//高精输入 {    set0(a);    char ch=getchar();    while (ch<‘0‘||ch>‘9‘)ch=getchar();      while (ch>=‘0‘&&ch<=‘9‘)      {          a.a[a.len++]=ch-‘0‘;        ch=getchar();    }    a.len--;      int change[mx+15];    for (int i=1;i<=a.len;i++)        change[i]=a.a[i];      for (int i=1;i<=a.len;i++)        a.a[i]=change[a.len-i+1];    while (a.a[a.len]==0)a.len--;}  inline void put(gaojing a)//高精输出{    for (int i=a.len;i>=1;i--)printf("%d",a.a[i]);    printf("\n");}inline bool operator < (const gaojing &a,const gaojing &b)//高精< {    if (a.len<b.len)return 1;    if (a.len>b.len)return 0;    for (int i=a.len;i>=1;i--)    {        if (a.a[i]<b.a[i])return 1;        if (a.a[i]>b.a[i])return 0;    }    return 0;}inline bool operator == (const gaojing &a,const gaojing &b)//高精=={    if (a.len!=b.len)return 0;    for (int i=a.len;i>=1;i--)    {        if (a.a[i]!=b.a[i])return 0;    }    return 1;}inline gaojing max(const gaojing &a,const gaojing &b)//高精max{    if (a<b)return b;    else return a;}inline gaojing min(const gaojing &a,const gaojing &b)//高精min{    if (a<b)return a;    else return b;} inline gaojing operator + (const gaojing &a,const gaojing &b)//高精+{    gaojing c;set0(c);      int maxlen=max(a.len,b.len);          for (int i=1;i<=maxlen;i++)          {              c.a[i]=c.a[i]+a.a[i]+b.a[i];              if (c.a[i]>=10)              {                  c.a[i+1]+=c.a[i]/10;                c.a[i]%=10;            }        }          c.len=maxlen+4;          while (!c.a[c.len]&&c.len>1) c.len--;        return c;}inline gaojing operator - (const gaojing &a,const gaojing &b)//高精-{    gaojing c;set0(c);    gaojing d;d=a;    for (int i=1;i<=b.len;i++)        {          c.a[i]=d.a[i]-b.a[i];          if (c.a[i]<0)          {              c.a[i]+=10;              int now=i+1;              while (!d.a[now])              {                  d.a[now]=9;                  now++;              }              d.a[now]--;          }        }    for (int i=b.len+1;i<=d.len;i++)c.a[i]=d.a[i];      c.len=d.len;      while (c.a[c.len]==0&&c.len>1)c.len--;    return c;}  inline gaojing operator * (const gaojing &a,const gaojing &b)//高精*{    gaojing c;set0(c);    for(int i=1;i<=a.len;i++)          for (int j=1;j<=b.len;j++)            c.a[i+j-1]+=a.a[i]*b.a[j];        c.len=a.len+b.len+5;      for (int i=1;i<=c.len;i++)          {            c.a[i+1]+=c.a[i]/10;            c.a[i]%=10;          }        while (!c.a[c.len]&&c.len>1)c.len--;    return c;}inline void div_by_2(gaojing &a){    for (int i=a.len;i>=1;i--)    {        if (a.a[i]&1 && i!=1)a.a[i-1]+=10;        a.a[i]/=2;    }    while (!a.a[a.len]&&a.len>1)a.len--;}inline gaojing operator / (gaojing a,const gaojing &b)//高精/{    gaojing l,r,ans;    set0(l);l.len=1;    set0(r);r=a;    set0(ans);ans.len=1;    while (l<r||l==r)    {        gaojing mid=l+r;        div_by_2(mid);        if(mid*b==a)return mid;        if(mid*b<a){ans=mid;l=mid+one;}        if(a<mid*b)r=mid-one;    }    return ans;}inline gaojing operator ^(const gaojing &a,int p)//高精^ {	gaojing ans=one,mult=a;	while (p)	{		if (p&1)ans=ans*mult;		mult=mult*mult;		p>>=1;	}	return ans;}inline void chushihua()//初始化,对0、1高精度常数赋值{    set0(zero); zero.len=1;    set0(one);one.len=1;one.a[1]=1;}int main(){	chushihua();}

  

 

……恩先这样吧还有的再加

写给自己看的模板