高精度加减乘-模板

新版

大数->超过整数型的数，很大的数

小数->int/long long范围的数

加法

#include <bits/stdc++.h>
using namespace std;

string x, y; 
vector<int> a, b, c;
void add(string x, string y)
{
	int t=0;
	for (int i=x.size()-1; i>=0; i--) a.push_back(x[i]-'0');
	for (int i=y.size()-1; i>=0; i--) b.push_back(y[i]-'0');
	
	for (int i=0; i<a.size() || i<b.size(); i++)
	{
		if (i<a.size()) t+=a[i];
		if (i<b.size()) t+=b[i];
		c.push_back(t%10);
		t/=10;
	}
	if (t) c.push_back(1);
}
int main()
{
    cin>>x>>y;
    add(x, y);
    
    for (int i=c.size()-1; i>=0; i--) printf("%d", c[i]);
    return 0;
}

减法

#include <bits/stdc++.h>
using namespace std;

string x, y; 
vector<int> a, b, c;
void sub(string x, string y)
{
	int t=0;
	for (int i=x.size()-1; i>=0; i--) a.push_back(x[i]-'0');
	for (int i=y.size()-1; i>=0; i--) b.push_back(y[i]-'0');
	
	for (int i=0; i<a.size(); i++)
	{
		t=a[i]-t;
		if (i<b.size()) t-=b[i];
		c.push_back((t+10)%10);
		
		if (t<0) t=1;
		else t=0;
	}
	while (c.size()>1 && c.back()==0) c.pop_back();
}
int main()
{
    cin>>x>>y;
    if (x.size()<y.size() || (x.size()==y.size() && x<y))
    {
    	swap(x, y);
    	printf("-");
	}
    sub(x, y);
    
    for (int i=c.size()-1; i>=0; i--) printf("%d", c[i]);
    return 0;
}

乘法，大数乘小数

#include <bits/stdc++.h>
using namespace std;

string x;
int b; 
vector<int> a, c;
void mul(string x, int b)
{
	int t=0;
	for (int i=x.size()-1; i>=0; i--) a.push_back(x[i]-'0');
	
	for (int i=0; i<a.size() || t; i++)
	{
		if (i<a.size()) t+=a[i]*b;
		c.push_back(t%10);
		t/=10;
	}
}
int main()
{
    cin>>x>>b;
    if (b==0)
    {
    	printf(0);
    	return 0;
	}
    mul(x, b);
    
    for (int i=c.size()-1; i>=0; i--) printf("%d", c[i]);
    return 0;
}

乘法，大数乘大数

#include <bits/stdc++.h>
using namespace std;

string x, y;
vector<int> a, b, c(100010);
void mul(string x, string y)
{
	int t=0;
	for (int i=x.size()-1; i>=0; i--) a.push_back(x[i]-'0');
	for (int i=y.size()-1; i>=0; i--) b.push_back(y[i]-'0');
	
	for (int i=0; i<a.size(); i++)
		for (int j=0; j<b.size(); j++)
		{
			c[i+j]+=a[i]*b[j]; //根据规律得来，观察上图可发现，C的下标是a的下标i+a的下标j，注意此处下标从0开始，图中下标从1开始，自己带入即可发现规律了
			c[i+j+1]+=c[i+j]/10;
			c[i+j]%=10;
		}
	
	while (c.size()>1 && c.back()==0) c.pop_back();
}
int main()
{
    cin>>x>>y;
    mul(x, y);
    
    for (int i=c.size()-1; i>=0; i--) printf("%d", c[i]);
    return 0;
}

除法，大数除以小数，洛谷的题开longlong才能过！！

#include <bits/stdc++.h>
using namespace std;

string x;
int b; 
vector<int> a, c;
void div(string x, int b, int &r)
{
	r=0;
	for (int i=x.size()-1; i>=0; i--) a.push_back(x[i]-'0');
	//正常情况顺着遍历即可，为了满足一些题目加减乘除混合，所以改变大数存储方式 
	for (int i=a.size()-1; i>=0; i--)
	{
		r=r*10+a[i];
		c.push_back(r/b);
		r%=b;
	}
	reverse(c.begin(), c.end());
	
	while (c.size()>1 && c.back()==0) c.pop_back();
}
int main()
{
    cin>>x>>b;
    if (b==0)
    {
    	printf(0);
    	return 0;
	}
	int r=0;
    div(x, b, r);
    
    for (int i=c.size()-1; i>=0; i--) printf("%d", c[i]);
    //printf("\n%d", r);  余数 
    return 0;
}

旧版

例题（一本通）：乘法 http://ybt.ssoier.cn:8088/problem_show.php?pid=1307

#include <bits/stdc++.h>
using namespace std;

struct bignum
{
    int len, s[1010];
}a, b;
string num1, num2;

void print(bignum x)
{
    //cout<<'('<<x.len<<')';
    int flag=0;
    for (int i=x.len-1; i>=0; i--)
    {
        if (x.s[i]!=0) flag=1;
        if (flag==1) cout<<x.s[i];
    }
    cout<<endl;
}
bignum cheng(bignum a, bignum b)
{
    bignum c;
    c.len=a.len+b.len;
    memset(c.s, 0, sizeof c.s);
    for (int i=0; i<b.len; i++)
        for (int j=0; j<a.len; j++)
            c.s[i+j]+=b.s[i]*a.s[j];

    for (int i=0; i<c.len; i++)
        if (c.s[i]>=10)
        {
            c.s[i+1]+=c.s[i]/10;
            c.s[i]%=10;
        }

    return c;
}
int main()
{
    cin>>num1>>num2;
    if (num1.size()<num2.size() || (num1.size()==num2.size() && num1<num2)) swap(num1, num2);
    a.len=num1.size(), b.len=num2.size();

    for (int i=0; i<num1.size(); i++)
        a.s[a.len-i-1]=num1[i]-'0';
    for (int i=0; i<num2.size(); i++)
        b.s[b.len-i-1]=num2[i]-'0';

    a=cheng(a, b);
    print(a);
    return 0;
}

http://ybt.ssoier.cn:8088/problem_show.php?pid=1174 （和前面一样的做法）

加法 http://ybt.ssoier.cn:8088/problem_show.php?pid=1168

#include <bits/stdc++.h>
using namespace std;

struct bignum
{
    int len, s[1010];
}a, b;
string num1, num2;

void print(bignum x)
{
    //cout<<'('<<x.len<<')';
    int flag=0;
    for (int i=x.len-1; i>=0; i--)
    {
        if (x.s[i]!=0) flag=1;
        if (flag==1) cout<<x.s[i];
    }
    cout<<endl;
}
bignum jia(bignum a, bignum b)
{
    bignum c;
    int len=0;
    if (a.len<b.len) swap(a, b);
    c.len=a.len, len=a.len;
    memset(c.s, 0, sizeof c.s);
    for (int i=0; i<len; i++)
        c.s[i]=a.s[i]+b.s[i];

    for (int i=0; i<len; i++)
        if (c.s[i]>=10)
        {
            c.s[i+1]++;
            c.s[i]-=10;
            if (i==len-1) c.len++;
        }
    return c;
}

int main()
{
    cin>>num1>>num2;
    if (num1.size()<num2.size() || (num1.size()==num2.size() && num1<num2)) swap(num1, num2);
    a.len=num1.size(), b.len=num2.size();

    for (int i=0; i<num1.size(); i++)
        a.s[a.len-i-1]=num1[i]-'0';
    for (int i=0; i<num2.size(); i++)
        b.s[b.len-i-1]=num2[i]-'0';

    a=jia(a, b);
    print(a);
    return 0;
}

　　减法 http://ybt.ssoier.cn:8088/problem_show.php?pid=1169

#include <bits/stdc++.h>
using namespace std;

struct bignum
{
    int len, s[1010];
}a, b;
string num1, num2;

void print(bignum x)
{
    //cout<<'('<<x.len<<')';
    int flag=0;
    for (int i=x.len-1; i>=0; i--)
    {
        if (x.s[i]!=0) flag=1;
        if (flag==1) cout<<x.s[i];
    }
    cout<<endl;
}
bignum jian(bignum a, bignum b)
{
    bignum c;
    int len=0;
    if (a.len<b.len) swap(a, b);
    c.len=a.len, len=a.len;
    memset(c.s, 0, sizeof c.s);
    for (int i=0; i<len; i++)
        c.s[i]=a.s[i]-b.s[i];

    for (int i=0; i<len; i++)
        if (c.s[i]<0)
        {
            c.s[i+1]--;
            c.s[i]+=10;
            if (i==len-1) c.len++;
        }
    return c;
}

int main()
{
    cin>>num1>>num2;
    if (num1.size()<num2.size() || (num1.size()==num2.size() && num1<num2)) swap(num1, num2);
    a.len=num1.size(), b.len=num2.size();

    for (int i=0; i<num1.size(); i++)
        a.s[a.len-i-1]=num1[i]-'0';
    for (int i=0; i<num2.size(); i++)
        b.s[b.len-i-1]=num2[i]-'0';

    a=jian(a, b);
    print(a);
    return 0;
}

扩展 1170：计算2的N次方 http://ybt.ssoier.cn:8088/problem_show.php?pid=1170

#include <bits/stdc++.h>
using namespace std;

struct bignum
{
    int len, s[1010];
}a, b;
int n;

void print(bignum x)
{
    //cout<<'('<<x.len<<')';
    int flag=0;
    for (int i=x.len-1; i>=0; i--)
    {
        if (x.s[i]!=0) flag=1;
        if (flag==1) cout<<x.s[i];
    }
    cout<<endl;
}
bignum cheng(bignum a, bignum b)
{
    bignum c;
    c.len=a.len+b.len;
    memset(c.s, 0, sizeof c.s);
    for (int i=0; i<b.len; i++)
        for (int j=0; j<a.len; j++)
            c.s[i+j]+=b.s[i]*a.s[j];

    for (int i=0; i<c.len; i++)
        if (c.s[i]>=10)
        {
            c.s[i+1]+=c.s[i]/10;
            c.s[i]%=10;
        }

    return c;
}
int main()
{
    cin>>n;
    b.s[0]=2, b.len=1;
    a.s[0]=1, a.len=1;
    for (int i=1; i<=n; i++)
        a=cheng(a, b);
    print(a);
    return 0;
}

1173：阶乘和（卡了最久） http://ybt.ssoier.cn:8088/problem_show.php?pid=1173 和洛谷的阶乘和一样 https://www.luogu.com.cn/problem/P1009

代码：

#include <bits/stdc++.h>
using namespace std;

struct bignum
{
    int len, s[1010];
}a, t, x;
int n;

void print(bignum x)
{
    //cout<<'('<<x.len<<')';
    int flag=0;
    for (int i=x.len-1; i>=0; i--)
    {
        if (x.s[i]!=0) flag=1;
        if (flag==1) cout<<x.s[i];
    }
    cout<<endl;
}
bignum jia(bignum a, bignum b)
{
    bignum c;
    int len=0;
    if (a.len<b.len) swap(a, b);
    c.len=a.len, len=a.len;
    memset(c.s, 0, sizeof c.s);
    for (int i=len-1; i>=0; i--)
        c.s[i]=a.s[i]+b.s[i];

    for (int i=0; i<len; i++)
        if (c.s[i]>=10)
        {
            c.s[i+1]++;
            c.s[i]-=10;
            if (i==len-1) c.len++;
        }
    return c;
}
bignum cheng(bignum a, bignum b)
{
    bignum c;
    c.len=a.len+b.len;
    memset(c.s, 0, sizeof c.s);
    for (int i=b.len-1; i>=0; i--)
        for (int j=a.len-1; j>=0; j--)
            c.s[i+j]+=b.s[i]*a.s[j];

    for (int i=0; i<c.len; i++)
        if (c.s[i]>=10)
        {
            c.s[i+1]+=c.s[i]/10;
            c.s[i]%=10;
        }

    return c;
}
int main()
{
    cin>>n;
    for (int i=1; i<=n; i++)
    {
        memset(t.s, 0, sizeof t.s);
        t.s[0]=1, t.len=1;
        for (int j=1; j<=i; j++)
        {
            memset(x.s, 0, sizeof x.s);
            if (j>=1 && j<=9) x.len=1, x.s[0]=j;
            else if (j>=10 && j<=99) x.len=2, x.s[1]=j/10%10, x.s[0]=j%10;
            else x.len=3, x.s[2]=j/100%10, x.s[1]=j/10%10, x.s[0]=j%10;

            t=cheng(t, x);
        }
        a=jia(a, t);
    }
    print(a);
    return 0;
}

posted @ 2023-10-28 22:42 cn是大帅哥886 阅读(44) 评论(0) 收藏举报

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