高精度加法
#include <bits/stdc++.h>
using namespace std;
const int L = 11010;
string add(string a, string b) //只限两个非负整数相加
{
string ans;
int na[L] = {0};
int nb[L] = {0};
int la = a.size();
int 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 = max(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;
cin >> a >> b;
cout << add(a, b);
return 0;
}
高精度减法
#include <bits/stdc++.h>
using namespace std;
const int L = 10010;
int na[L] = {0};
int nb[L] = {0};
string sub(string a, string b)
{
string ans;
int la = a.size();
int lb = b.size();
int lmax = max(la, lb);
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';
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;
cin >> a >> b;
int la = a.size();
int lb = b.size();
if(lb>la)
{
swap(a,b);
cout<<"-"<<sub(a,b);
return 0;
}
cout << sub(a,b) ;
return 0;
}
高精度乘法
#include <bits/stdc++.h>
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;
cin >> a >> b;
cout << mul(a, b);
return 0;
}
高精度乘法[fft优化]
#include <bits/stdc++.h>
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()
{
string a, b;
cin >> a >> b;
cout << mul(a, b);
return 0;
}
高精度除法(附带取模)
#include <bits/stdc++.h>
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;
cin >> a >> b;
cout << div(a, b, 1); //将1改成2可将高精度除法改为高精度取模
return 0;
}
高精度阶乘
#include <bits/stdc++.h>
using namespace std;
const int L = 100005;
int a[L];
string fac(int n)
{
string ans;
if (n == 0) return "1";
fill(a, a + L, 0);
int s = 0, m = n;
while (m) a[++s] = m % 10, m /= 10;
for (int i = n - 1; i >= 2; i--)
{
int w = 0;
for (int j = 1; j <= s; j++) a[j] = a[j] * i + w, w = a[j] / 10, a[j] = a[j] % 10;
while (w) a[++s] = w % 10, w /= 10;
}
while (!a[s]) s--;
while (s >= 1) ans += a[s--] + '0';
return ans;
}
int main()
{
int n;
cin >> n;
cout << fac(n);
return 0;
}
高精度幂
#include <bits/stdc++.h>
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;
}
string Pow(string a, int n)
{
if (n == 1) return a;
if (n & 1) return mul(Pow(a, n - 1), a);
string ans = Pow(a, n / 2);
return mul(ans, ans);
}
int main()
{
cin.sync_with_stdio(false);
string a;
int b;
cin >> a >> b;
cout << Pow(a, b);
return 0;
}
高精度GCD
#include <bits/stdc++.h>
using namespace std;
const int L=110;
string add(string a,string b)
{
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;
}
string mul(string a,string b)
{
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 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;
}
bool judge(string s)//判断s是否为全0串
{
for(int i=0;i<s.size();i++)
if(s[i]!='0') return false;
return true;
}
string gcd(string a,string b)//求最大公约数
{
string t;
while(!judge(b))//如果余数不为0，继续除
{
t=a;//保存被除数的值
a=b;//用除数替换被除数
b=div(t,b,2);//用余数替换除数
}
return a;
}
int main()
{
string a,b;
cin>>a>>b;
cout<<gcd(a,b);
return 0;
}