高精度汇总

1 高精度加法

#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
using namespace std;
char s1[205],s2[205];
int sa[205],sb[205];
int len1,len2,lenn,lenm;
int ans[415]; 
int main(){
	scanf("%s
%s",s1+1,s2+1);
	len1=strlen(s1+1);
	len2=strlen(s2+1);
	for(int i=1,j=len1;i<=len1;i++,j--){
		sa[j]=s1[i]-'0';
	}
	for(int i=1,j=len2;i<=len2;i++,j--){
		sb[j]=s2[i]-'0';
	}
	lenn=len1;lenm=len2;
	int lenx=max(lenn,lenm)+1;
	for(int i=lenx;i>=1;i--){
		ans[i]=sa[i]+sb[i];
	}
	for(int i=1;i<=lenx;i++){
		if(ans[i]>=10){
			ans[i+1]++;
			ans[i]%=10;
		}
	}
	while(ans[lenx]==0&&lenx>1) lenx--;
	for(int i=lenx;i>=1;i--){
		printf("%d",ans[i]);
	}
	return 0;
}

2 高精度减法

有符号判断

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std;
char num1[50005],num2[50005];
int numa1[50005],numa2[50005],ans[50005];
int len1,len2,lenn,lenm;
int main( ){
	scanf("%s%s",num1+1,num2+1);
	memset(ans,0,sizeof(ans));
	len1=strlen(num1+1);
	len2=strlen(num2+1);
	lenm=max(len1,len2);
	bool big=0;//big=1 :1比2大  big=0 :2比1大 
	if(len1!=len2) big=(len1>len2);
	else{
		for(int i=1;i<=len1;i++){
			if(num1[i]>num2[i]){
				big=1;
				break;
			}
			if(num1[i]<num2[i]){
				big=0;
				break;
			}
		}
	} 
	if(big==0){
		swap(len1,len2);
		swap(num1,num2);
	}
	for(int i=1,j=len1;i<=len1;i++,j--) numa1[i]=num1[j]-'0';
	for(int i=1,j=len2;i<=len2;i++,j--) numa2[i]=num2[j]-'0';
	for(int i=1;i<=lenm;i++){
		ans[i]=numa1[i]-numa2[i];
	}
	for(int i=1;i<=lenm;i++){
		if(ans[i]<0){
			ans[i]+=10;
			ans[i+1]--;
		}
		else continue;
	}
	lenn=lenm;
	while(ans[lenn]==0&&lenn>=0){
		lenn--;
		if(lenn==-1) {
			printf("0");
			return 0;
		}
	}
	if(big==0) printf("-");
	for(int i=lenn;i>=1;i--){
		printf("%d",ans[i]);
	}
	return 0;
} 

3 高精度除以低精度

A 本人写法

//本人写法 
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
char num1[205];
int ans[205],numa1[205],numa2[205]; 
int len1,len2,lenm;
int ys;//余数 
int ss;//商 
int f=0;
long long x;
int main( ){
	scanf("%s",num1+1);
	len1=strlen(num1+1);
	scanf("%lld",&x);
	if(x==0) {
		printf("0");
		return 0;
	}
	for(int i=1;i<=len1;i++)
	{
		numa1[i]=num1[i]-'0';
	 } 
	for(int i=1;i<=len1;i++){
		ys=numa1[i]%x;
		ss=numa1[i]/x;
		numa1[i]=ss;
		numa1[i+1]+=ys*10;
	}
	for(int i=1;i<=len1;i++){
		if(f==0&&numa1[i]==0) printf("0");
		if(f==0&&numa1[i]>0) f=1;
		if(f==1){
			printf("%d",numa1[i]);
		}
		
	}
	return 0;
} 

B 洛谷题解其一

#include<cstdio>
#include <cstring>
using namespace std;
int a[10000];
int y; 
int l=0; 
int yushu; 
int shang;
int f=0; 
int main()
{
    char c=getchar();
    while (c>='0' && c<='9')
    {
        l++;
        a[l]=(c-'0');
        c=getchar();    
    }
    //scanf("%s",a);
    //l=strlen(a);
    scanf("%d",&y); 
    for (int i=1;i<=l;i++) 
    {
        yushu=a[i]%y;//第i位的余数 
        shang=a[i]/y;//第i位的商 
        a[i]=shang;
        a[i+1]+=yushu*10;//把余数弄到下一位
    }
    for (int i=1;i<=l;i++)
    {
        if (f==0 && a[i]>0) f=1;
        if (f==1) printf("%d",a[i]);
    }
    return 0; 
}

4 高精度乘法

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
char num1[2005],num2[2005];
int numa1[2005],numa2[2005],ans[4005];
int len1,len2,lenn,lenm;
int lenn1,lenn2;
int main( ){
	scanf("%s%s",num1+1,num2+1);
	len1=strlen(num1+1);
	len2=strlen(num2+1);
	lenn1=len1;
	lenn2=len2;
	memset(numa1,0,sizeof(numa1));
	memset(numa2,0,sizeof(numa2));
	memset(ans,0,sizeof(ans));
	for(int i=1,j=len1;i<=len1;i++,j--) numa1[i]=num1[j]-'0';
	for(int i=1,j=len2;i<=len2;i++,j--) numa2[i]=num2[j]-'0';
	//for(int i=lenn1;i>=1;i--){
	//	while(numa1[i]==0) len1--;
	//}
	//for(int i=lenn2;i>=1;i--){
	//	while(numa2[i]==0) len2--;
	//}
	//前导0考虑 //判断0的情况
	for(int i=1;i<=len1;i++){
		for(int j=1;j<=len2;j++){
			ans[i+j-1]+=numa1[i]*numa2[j];
			if(ans[i+j-1]>=10){
				ans[i+j]+=ans[i+j-1]/10;
				ans[i+j-1]%=10;	
			}
		}
	} 
	lenn=len1+len2;
	for(int i=1;i<=lenn;i++){
		if(ans[i]>=10){
			ans[i+1]+=ans[i]/10;
			ans[i]%=10;
		}
	}
	while(ans[lenn]==0&&lenn>1){
		lenn--;
	}
	for(int i=lenn;i>=1;i--){
		printf("%d",ans[i]);
	}
	return 0;
}

5 高精度(2^N)

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream> 
#include <iomanip>
using namespace std;
int ans[1005]={0};
void doubl2(int c[]){
	for(int i=1;i<=c[0];i++){
		c[i]=c[i]*2;
	}
	for(int i=1;i<=c[0];i++){
		if(c[i]>=10){
			c[i+1]+=ans[i]/10;//ans??
			c[i]%=10; 
		}
	}	
	if(c[c[0]+1]) c[0]++; 
}
void prta(int c[]){
	for(int i=1;i<=c[0];i++){
		printf("%d",c[c[0]+1-i]);
	}
}
int main( ){
	ans[1]=1;
	ans[0]=1;
	int n;
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
		doubl2(ans);
	}
	prta(ans);
	return 0;
}

6 高精度合集

By _cyl

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
typedef long long ll;
const int N=2e4+5;
char s[N];
ll qwq,MOD;
struct bignum
{
    int a[N],len;
    bignum(){memset(a,0,sizeof(a));len=0;}
    void push()
    {
        for(int i=0;i<=len;i++)
        {
            if(a[i]>=10) 
            {
                if(i==len) len++;
                a[i+1]+=a[i]/10;
                a[i]%=10;
            }
        }
    }
  void read()
    {
        scanf("%s",s);
        len=strlen(s)-1;
        for(int i=0,j=len;i<=len;i++,j--) a[i]=s[j]-'0';
    }
    void write()
    {
        for(int i=len;i>=0;i--) printf("%d",a[i]);
        putchar('
'); 
    }
  friend inline bool operator <(const bignum &x,const bignum &y)
  {
    if(x.len!=y.len) return x.len<y.len;
    for(int i=x.len;i>=0;i--) if(x.a[i]!=y.a[i]) return x.a[i]<y.a[i];
  	return 0; 
  }
    friend inline bignum operator +(const bignum &x,const bignum &y)
    {
        bignum ans; 
        ans.len=max(x.len,y.len);
        for(int i=0;i<=ans.len;i++) ans.a[i]=x.a[i]+y.a[i];
        ans.push();
        return ans;
    }
  friend inline bignum operator -(const bignum &fi,const bignum &se)
  {
    bignum ans,x,y;
    int op=1;
    if(fi<se) op=-1,x=se,y=fi;
    else
      {
	if(!(se<fi)) return ans;
	x=fi,y=se;
      }
    //x.write();y.write();
    ans.len=x.len;
    for(int i=0;i<=ans.len;i++)
      {
	ans.a[i]=x.a[i]-y.a[i];
	if(ans.a[i]<0) ans.a[i]+=10,x.a[i+1]--;
      }
    for(int i=ans.len;i>=0;i--) if(ans.a[i]==0) ans.len--;else break;
    ans.a[ans.len]*=op;
    return ans;
  }
    friend inline bignum operator *(const bignum &x,const bignum &y)
    {
        bignum ans;
        if((x.len==0 && x.a[0]==0) || (y.len==0 && y.a[0]==0))return ans;
        ans.len=x.len+y.len;
        for(int i=0;i<=x.len;i++)
            for(int j=0;j<=y.len;j++)
                ans.a[i+j]+=x.a[i]*y.a[j];
        ans.push();
        return ans;
    }
  friend inline bignum operator /(const bignum &x,const ll &y)
  {
    ll alpha=0;
    bignum ans;
    for(int i=x.len;i>=0;i--)
      {
	alpha=alpha*10+x.a[i];
	if(alpha>=y && ans.len==0) ans.len=i;
	ans.a[i]=alpha/y;alpha%=y;
      }
    MOD=alpha;
    return ans;
  }
}f[2];
int main()
{
  //f[0].read();//f[0].write();
  //scanf("%lld",&qwq);//f[1].write();
  //(f[0]/qwq).write();
  //printf("%lld
",MOD);
  f[0].read();f[1].read();(f[0]-f[1]).write();
  return 0;
}