高精度模板

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转自 http://www.mamicode.com/info-detail-454902.html
1.加法

#include<iostream>#include<cstring>#include<algorithm>using namespace std;const int L=110;string add(string a,string b)//只限两个非负整数相加{    string ans;    int na[L]={0},nb[L]={0};    int la=a.size(),lb=b.size();    for(int i=0;i<la;i++) na[la-1-i]=a[i]-'0';    for(int i=0;i<lb;i++) nb[lb-1-i]=b[i]-'0';    int lmax=la>lb?la:lb;    for(int i=0;i<lmax;i++) na[i]+=nb[i],na[i+1]+=na[i]/10,na[i]%=10;    if(na[lmax]) lmax++;    for(int i=lmax-1;i>=0;i--) ans+=na[i]+'0';    return ans;}int main(){    string a,b;    while(cin>>a>>b) cout<<add(a,b)<<endl;    return 0;}

2 减法

#include<iostream>#include<cstring>#include<algorithm>using namespace std;const int L=110;string sub(string a,string b)//只限大的非负整数减小的非负整数{    string ans;    int na[L]={0},nb[L]={0};    int la=a.size(),lb=b.size();    for(int i=0;i<la;i++) na[la-1-i]=a[i]-'0';    for(int i=0;i<lb;i++) nb[lb-1-i]=b[i]-'0';    int lmax=la>lb?la:lb;    for(int i=0;i<lmax;i++)    {        na[i]-=nb[i];        if(na[i]<0) na[i]+=10,na[i+1]--;    }    while(!na[--lmax]&&lmax>0)  ;lmax++;    for(int i=lmax-1;i>=0;i--) ans+=na[i]+'0';    return ans;}int main(){    string a,b;    while(cin>>a>>b) cout<<sub(a,b)<<endl;    return 0;}

3.乘法

#include<iostream>#include<cstring>#include<algorithm>using namespace std;const int L=110;string mul(string a,string b)//高精度乘法a,b,均为非负整数{    string s;    int na[L],nb[L],nc[L],La=a.size(),Lb=b.size();//na存储被乘数，nb存储乘数，nc存储积    fill(na,na+L,0);fill(nb,nb+L,0);fill(nc,nc+L,0);//将na,nb,nc都置为0    for(int i=La-1;i>=0;i--) na[La-i]=a[i]-'0';//将字符串表示的大整形数转成i整形数组表示的大整形数    for(int i=Lb-1;i>=0;i--) nb[Lb-i]=b[i]-'0';    for(int i=1;i<=La;i++)        for(int j=1;j<=Lb;j++)        nc[i+j-1]+=na[i]*nb[j];//a的第i位乘以b的第j位为积的第i+j-1位（先不考虑进位）    for(int i=1;i<=La+Lb;i++)        nc[i+1]+=nc[i]/10,nc[i]%=10;//统一处理进位    if(nc[La+Lb]) s+=nc[La+Lb]+'0';//判断第i+j位上的数字是不是0    for(int i=La+Lb-1;i>=1;i--)        s+=nc[i]+'0';//将整形数组转成字符串    return s;}int main(){    string a,b;    while(cin>>a>>b) cout<<mul(a,b)<<endl;    return 0;}

4.高精度乘高精度FFT优化算法

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <cmath>#include <map>#include <queue>#include <set>#include <vector>using namespace std;#define L(x) (1 << (x))const double PI = acos(-1.0);const int Maxn = 133015;double ax[Maxn], ay[Maxn], bx[Maxn], by[Maxn];char sa[Maxn/2],sb[Maxn/2];int sum[Maxn];int x1[Maxn],x2[Maxn];int revv(int x, int bits){    int ret = 0;    for (int i = 0; i < bits; i++)    {        ret <<= 1;        ret |= x & 1;        x >>= 1;    }    return ret;}void fft(double * a, double * b, int n, bool rev){    int bits = 0;    while (1 << bits < n) ++bits;    for (int i = 0; i < n; i++)    {        int j = revv(i, bits);        if (i < j)            swap(a[i], a[j]), swap(b[i], b[j]);    }    for (int len = 2; len <= n; len <<= 1)    {        int half = len >> 1;        double wmx = cos(2 * PI / len), wmy = sin(2 * PI / len);        if (rev) wmy = -wmy;        for (int i = 0; i < n; i += len)        {            double wx = 1, wy = 0;            for (int j = 0; j < half; j++)            {                double cx = a[i + j], cy = b[i + j];                double dx = a[i + j + half], dy = b[i + j + half];                double ex = dx * wx - dy * wy, ey = dx * wy + dy * wx;                a[i + j] = cx + ex, b[i + j] = cy + ey;                a[i + j + half] = cx - ex, b[i + j + half] = cy - ey;                double wnx = wx * wmx - wy * wmy, wny = wx * wmy + wy * wmx;                wx = wnx, wy = wny;            }        }    }    if (rev)    {        for (int i = 0; i < n; i++)            a[i] /= n, b[i] /= n;    }}int solve(int a[],int na,int b[],int nb,int ans[]){    int len = max(na, nb), ln;    for(ln=0; L(ln)<len; ++ln);    len=L(++ln);    for (int i = 0; i < len ; ++i)    {        if (i >= na) ax[i] = 0, ay[i] =0;        else ax[i] = a[i], ay[i] = 0;    }    fft(ax, ay, len, 0);    for (int i = 0; i < len; ++i)    {        if (i >= nb) bx[i] = 0, by[i] = 0;        else bx[i] = b[i], by[i] = 0;    }    fft(bx, by, len, 0);    for (int i = 0; i < len; ++i)    {        double cx = ax[i] * bx[i] - ay[i] * by[i];        double cy = ax[i] * by[i] + ay[i] * bx[i];        ax[i] = cx, ay[i] = cy;    }    fft(ax, ay, len, 1);    for (int i = 0; i < len; ++i)        ans[i] = (int)(ax[i] + 0.5);    return len;}string mul(string sa,string sb){    int l1,l2,l;    int i;    string ans;    memset(sum, 0, sizeof(sum));    l1 = sa.size();    l2 = sb.size();    for(i = 0; i < l1; i++)        x1[i] = sa[l1 - i - 1]-'0';    for(i = 0; i < l2; i++)        x2[i] = sb[l2-i-1]-'0';    l = solve(x1, l1, x2, l2, sum);    for(i = 0; i<l || sum[i] >= 10; i++) // 进位    {        sum[i + 1] += sum[i] / 10;        sum[i] %= 10;    }    l = i;    while(sum[l] <= 0 && l>0)    l--; // 检索最高位    for(i = l; i >= 0; i--)    ans+=sum[i] + '0'; // 倒序输出    return ans;}int main(){    cin.sync_with_stdio(false);    string a,b;    while(cin>>a>>b) cout<<mul(a,b)<<endl;    return 0;}

5.除法

#include<iostream>#include<cstring>#include<algorithm>using namespace std;const int L=110;int sub(int *a,int *b,int La,int Lb){    if(La<Lb) return -1;//如果a小于b，则返回-1    if(La==Lb)    {        for(int i=La-1;i>=0;i--)            if(a[i]>b[i]) break;            else if(a[i]<b[i]) return -1;//如果a小于b，则返回-1    }    for(int i=0;i<La;i++)//高精度减法    {        a[i]-=b[i];        if(a[i]<0) a[i]+=10,a[i+1]--;    }    for(int i=La-1;i>=0;i--)        if(a[i]) return i+1;//返回差的位数    return 0;//返回差的位数}string div(string n1,string n2,int nn)//n1,n2是字符串表示的被除数，除数,nn是选择返回商还是余数{    string s,v;//s存商,v存余数     int a[L],b[L],r[L],La=n1.size(),Lb=n2.size(),i,tp=La;//a，b是整形数组表示被除数，除数，tp保存被除数的长度     fill(a,a+L,0);fill(b,b+L,0);fill(r,r+L,0);//数组元素都置为0     for(i=La-1;i>=0;i--) a[La-1-i]=n1[i]-'0';     for(i=Lb-1;i>=0;i--) b[Lb-1-i]=n2[i]-'0';     if(La<Lb || (La==Lb && n1<n2)) {            //cout<<0<<endl;     return n1;}//如果a<b,则商为0，余数为被除数     int t=La-Lb;//除被数和除数的位数之差     for(int i=La-1;i>=0;i--)//将除数扩大10^t倍        if(i>=t) b[i]=b[i-t];        else b[i]=0;     Lb=La;     for(int j=0;j<=t;j++)     {         int temp;         while((temp=sub(a,b+j,La,Lb-j))>=0)//如果被除数比除数大继续减         {             La=temp;             r[t-j]++;         }     }     for(i=0;i<L-10;i++) r[i+1]+=r[i]/10,r[i]%=10;//统一处理进位     while(!r[i]) i--;//将整形数组表示的商转化成字符串表示的     while(i>=0) s+=r[i--]+'0';     //cout<<s<<endl;     i=tp;     while(!a[i]) i--;//将整形数组表示的余数转化成字符串表示的</span>     while(i>=0) v+=a[i--]+'0';     if(v.empty()) v="0";     //cout<<v<<endl;     if(nn==1) return s;     if(nn==2) return v;}int main(){    string a,b;    while(cin>>a>>b) cout<<div(a,b,1)<<endl;    return 0;}

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