大数加法和大数乘法模板函数

好多函数经常用到,于是写好模板以后直接调用~~~

大数加法:

//Must so
#include<iostream>
#include<algorithm>
#include<string>
using namespace std;

string Bigadd(string s1,string s2)
{
     string s;
     int l1 =  s1.size(),l2 = s2.length(),mx;
	if (l1 > l2)
	{
		mx = l1;
		string ex(l1 - l2,'0');
		s2 = ex + s2;
	}
	else
	{
		mx = l2;
		string ex(l2 - l1,'0');
		s1 = ex + s1;
	}
	int jingwei = 0,ans;
	for (int i = mx - 1;i >= 0;i--)
	{
		ans = s1[i] + s2[i] - 96 + jingwei;
		jingwei = ans / 10;
		char a = ans % 10 + 48;
		s = a + s;
	}
	char a = jingwei + 48;
	if (a != '0') s = a + s;
	return s;
}
int main()
{
    string a,b;
    cin>>a>>b;
    cout<<Bigadd(a,b)<<endl;
    return 0;
}

<span style="font-family: Arial, Helvetica, sans-serif; background-color: rgb(255, 255, 255);">
</span>

<span style="font-family: Arial, Helvetica, sans-serif; background-color: rgb(255, 255, 255);">关于大数加法,我记得见过一种非常牛逼的处理,当时大致估摸最后的结果不超过10的32次方,于是某位大神用2个LL,一个记录尾,一个记录头,模拟了大数加法的一部分,超级厉害!!!</span>

相关题目和代码最后贴上供参考~~~

大数乘法:

<span style="font-family:Arial, Helvetica, sans-serif;"><span style="background-color: rgb(255, 255, 255);"></span></span><pre name="code" class="cpp">//Must so
#include<iostream>
#include<algorithm>
#include<string>
#include<cstring>
#include<ctype.h>
#include<queue>
#include<vector>
#include<set>
#include<cstdio>
#include<cmath>
#define mem(a,x) memset(a,x,sizeof(a))
#define ex(x) ((x)*(x))
#define inf 1<<29
#define NN 1000006
using namespace std;
const double PI = acos(-1.0);
typedef long long LL;
int c[10000];
string bigmul(string a,string b)
{
    int aa = a.size(),bb = b.size(),cc = aa + bb;
    for (int i = 0;i < cc;i++) c[i] = 0;
    reverse(a.begin(),a.end());
    reverse(b.begin(),b.end());
    for (int i = 0; i < aa; i++)
    {
        for (int j = 0; j < bb; j++)
        {
            c[i+j] += (a[i]-48)*(b[j]-48);
        }
    }
    for (int i = 0; i < cc-1; i++)
    {
        c[i+1] += c[i]/10;
        c[i] = c[i]%10;
    }
    int i = cc-1;
    while (c[i--] == 0);
    string ans;
    for (int j = i+1;j >= 0;j--) ans += (c[j]+48);
    return ans;
}
int main()
{
    string a,b;
    while (cin>>a>>b)
    {
        cout<<bigmul(a,b)<<endl;
    }
    return 0;
}

关于大数加法一个巧妙的方法:

题目链接:点击打开链接

这个题目我是用母函数做的,大牛们则是完全背包做的....我是很久之后才被告知母函数是冷门技巧....

AC代码:

//Must so
//母函数
#include<cstdio>
#include<iostream>
using namespace std;
long long c1[1001],c2[1001];
long long pre1[1001],pre2[1001];//c1,c2是尾数，pre1,pre2是前面的高位部分
int main()
{
    long long mod = 1;
    for (int i = 0;i < 18;i++) mod *= 10;//18位
    int n,k;
    while (~scanf("%d%d",&n,&k))
    {
        for (int i = 0;i <= n;i++)
        {
            c1[i] = 1;
            pre1[i] = 0;
            c2[i] = 0;
            pre2[i] = 0;
        }
        for (int i = 2;i <= k;i++)
        {
            for (int j = 0;j <= n;j++)
            {
                for (int k = 0;k+j <= n;k += i)
                {
                    c2[j+k] += c1[j];
                    pre2[j+k] += pre1[j];
                   // if (pre2[j+k]) printf("---%lld---???\n",pre2[j+k]);
                    pre2[j+k] += c2[j+k]/mod;//进位
                    c2[j+k] %= mod;//取余

                }
            }
            for (int j = 0;j <= n;j++)
            {
                c1[j] = c2[j];
                pre1[j] = pre2[j];
                c2[j] = 0;
                pre2[j] = 0;
            }
        }
       // printf("-----%lld-----\n",pre1[n]);
        if (pre1[n]) printf("%lld",pre1[n]);
        printf("%lld\n",c1[n]);
    }
    return 0;
}

</pre><pre>