简单版高精度
#include <bits/stdc++.h>
using namespace std;
struct bigint {
int d[1000];
int len;
bigint() : len(0) {
memset(d, 0, sizeof(d));
}
void fix() {
while (len >= 2 && d[len - 1] == 0) len--;
}
bool operator < (const bigint &b) const {
if (len != b.len) return len < b.len;
for (int i = len - 1; i >= 0; i++) {
if (d[i] != b.d[i]) return d[i] < b.d[i];
}
return 0;
}
};
bigint change(string s) {
bigint a;
a.len = (int)s.length();
for (int i = 0; i < a.len; i++){
a.d[i] = s[a.len - i - 1] - '0';
}
return a;
}
bigint change(long long x) {
string s = to_string(x);
return change(s);
}
bigint operator + (const bigint &a, const bigint &b) {
bigint c;
int carry = 0;
for (int i = 0; i < a.len || i < b.len; i++) {
int tmp = a.d[i] + b.d[i] + carry;
c.d[c.len++] = tmp % 10;
carry = tmp / 10;
}
if (carry) c.d[c.len++] = carry;
return c;
}
bigint operator - (const bigint &a, const bigint &b) {
bigint c = a;
for (int i = 0; i <= b.len; i++) {
if (i != b.len) c.d[i] -= b.d[i];
while (c.d[i] < 0) {
c.d[i] += 10; c.d[i + 1]--;
}
}
c.fix();
return c;
}
bigint operator * (const bigint &a, int b) {
bigint c;
int carry = 0;
for (int i = 0; i < a.len; i++) {
int tmp = a.d[i] * b + carry;
c.d[c.len++] = tmp % 10;
carry = tmp / 10;
}
while (carry != 0) {
c.d[c.len++] = carry % 10; carry /= 10;
}
return c;
}
bigint operator * (const bigint &a, const bigint &b) {
bigint c; c.len = a.len + b.len;
for (int i = 0; i < a.len; i++) {
for (int j = 0; j < b.len; j++) {
c.d[i + j] += a.d[i] * b.d[j];
}
}
for (int i = 0; i < c.len - 1; i++) {
c.d[i + 1] += c.d[i] / 10; c.d[i] %= 10;
}
c.fix();
return c;
}
bigint divide(bigint a, int b, int &r) {
bigint c;
c.len = a.len;
for (int i = a.len - 1; i >= 0; i--) {
r = r * 10 + a.d[i];
if (r < b) c.d[i] = 0;
else {
c.d[i] = r / b; r %= b;
}
}
c.fix();
return c;
}
void print(bigint a) {
for (int i = a.len - 1; i >= 0; i--) {
printf("%d",a.d[i]);
}
puts("");
}
int main() {
bigint now = change("2");
print(now);
for (int i = 1; i <= 100; i++) {
now = now * 2;
print(now);
}
return 0;
}
完全版高精度
#include <bits/stdc++.h>
using namespace std;
constexpr int base = 1000000000;
constexpr int base_digits = 9;
struct bigint {
// value == 0 is represented by empty z
vector<int> z; // digits
// sign == 1 <==> value >= 0
// sign == -1 <==> value < 0
int sign;
bigint() : sign(1) {}
bigint(long long v) { *this = v; }
bigint& operator=(long long v) {
sign = v < 0 ? -1 : 1;
v *= sign;
z.clear();
for (; v > 0; v = v / base) z.push_back((int)(v % base));
return *this;
}
bigint(const string& s) { read(s); }
void trim() {
while (!z.empty() && z.back() == 0) z.pop_back();
if (z.empty()) sign = 1;
}
void read(const string& s) {
sign = 1;
z.clear();
int pos = 0;
while (pos < s.size() && (s[pos] == '-' || s[pos] == '+')) {
if (s[pos] == '-') sign = -sign;
++pos;
}
for (int i = (int)s.size() - 1; i >= pos; i -= base_digits) {
int x = 0;
for (int j = max(pos, i - base_digits + 1); j <= i; j++)
x = x * 10 + s[j] - '0';
z.push_back(x);
}
trim();
}
friend istream& operator>>(istream& stream, bigint& v) {
string s;
stream >> s;
v.read(s);
return stream;
}
friend ostream& operator<<(ostream& stream, const bigint& v) {
if (v.sign == -1) stream << '-';
stream << (v.z.empty() ? 0 : v.z.back());
for (int i = (int)v.z.size() - 2; i >= 0; --i)
stream << setw(base_digits) << setfill('0') << v.z[i];
return stream;
}
/*以上为读入和输出高精度的必要方法*/
long long longValue() const {
long long res = 0;
for (int i = (int)z.size() - 1; i >= 0; i--) res = res * base + z[i];
return res * sign;
}
/*特定时候可以用来加速*/
bigint& operator*=(int v) {
if (v < 0) sign = -sign, v = -v;
for (int i = 0, carry = 0; i < z.size() || carry; i++) {
if (i == z.size()) z.push_back(0);
long long cur = (long long)z[i] * v + carry;
carry = (int)(cur / base);
z[i] = (int)(cur % base);
}
trim();
return *this;
}
bigint operator*(int v) const { return bigint(*this) *= v; }
bigint& operator/=(int v) {
if (v < 0) sign = -sign, v = -v;
for (int i = (int)z.size() - 1, rem = 0; i >= 0; --i) {
long long cur = z[i] + rem * (long long)base;
z[i] = (int)(cur / v);
rem = (int)(cur % v);
}
trim();
return *this;
}
bigint operator/(int v) const { return bigint(*this) /= v; }
int operator%(int v) const {
if (v < 0) v = -v;
int m = 0;
for (int i = (int)z.size() - 1; i >= 0; --i)
m = (int)((z[i] + m * (long long)base) % v);
return m * sign;
}
/*以上为高精度和整数的乘法除法取模,没有依赖项,互相之间也不依赖*/
bool operator<(const bigint& v) const {
if (sign != v.sign) return sign < v.sign;
if (z.size() != v.z.size())
return z.size() * sign < v.z.size() * v.sign;
for (int i = (int)z.size() - 1; i >= 0; i--)
if (z[i] != v.z[i])
return z[i] * sign < v.z[i] * sign;
return false;
}
bool operator<=(const bigint& v) const { return !(v < *this); }
bool operator>=(const bigint& v) const { return !(*this < v); }
bool operator!=(const bigint& v) const { return *this < v || v < *this; }
friend bigint operator-(bigint v) {
if (!v.z.empty()) v.sign = -v.sign;
return v;
}
bigint& operator+=(const bigint& other) {
if (sign == other.sign) {
for (int i = 0, carry = 0; i < other.z.size() || carry; ++i) {
if (i == z.size()) z.push_back(0);
z[i] += carry + (i < other.z.size() ? other.z[i] : 0);
carry = z[i] >= base;
if (carry) z[i] -= base;
}
}
else if (other != 0 /* prevent infinite loop */) {
*this -= -other;
}
return *this;
}
friend bigint operator+(bigint a, const bigint& b) { return a += b; }
bigint& operator-=(const bigint& other) {
if (sign == other.sign) {
if ((sign == 1 && *this >= other) || (sign == -1 && *this <= other)) {
for (int i = 0, carry = 0; i < other.z.size() || carry; ++i) {
z[i] -= carry + (i < other.z.size() ? other.z[i] : 0);
carry = z[i] < 0;
if (carry) z[i] += base;
}
trim();
}
else {
*this = other - *this;
this->sign = -this->sign;
}
}
else {
*this += -other;
}
return *this;
}
friend bigint operator-(bigint a, const bigint& b) { return a -= b; }
/*以上为高精度加减所依赖项*/
static vector<int> convert_base(const vector<int>& a, int old_digits, int new_digits) {
vector<long long> p(max(old_digits, new_digits) + 1);
p[0] = 1;
for (int i = 1; i < p.size(); i++) p[i] = p[i - 1] * 10;
vector<int> res;
long long cur = 0;
int cur_digits = 0;
for (int v : a) {
cur += v * p[cur_digits];
cur_digits += old_digits;
while (cur_digits >= new_digits) {
res.push_back(int(cur % p[new_digits]));
cur /= p[new_digits];
cur_digits -= new_digits;
}
}
res.push_back((int)cur);
while (!res.empty() && res.back() == 0) res.pop_back();
return res;
}
typedef vector<long long> vll;
static vll karatsubaMultiply(const vll& a, const vll& b) {
int n = (int)a.size();
vll res(n + n);
if (n <= 32) {
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++) res[i + j] += a[i] * b[j];
return res;
}
int k = n >> 1;
vll a1(a.begin(), a.begin() + k);
vll a2(a.begin() + k, a.end());
vll b1(b.begin(), b.begin() + k);
vll b2(b.begin() + k, b.end());
vll a1b1 = karatsubaMultiply(a1, b1);
vll a2b2 = karatsubaMultiply(a2, b2);
for (int i = 0; i < k; i++) a2[i] += a1[i];
for (int i = 0; i < k; i++) b2[i] += b1[i];
vll r = karatsubaMultiply(a2, b2);
for (int i = 0; i < a1b1.size(); i++) r[i] -= a1b1[i];
for (int i = 0; i < a2b2.size(); i++) r[i] -= a2b2[i];
for (int i = 0; i < r.size(); i++) res[i + k] += r[i];
for (int i = 0; i < a1b1.size(); i++) res[i] += a1b1[i];
for (int i = 0; i < a2b2.size(); i++) res[i + n] += a2b2[i];
return res;
}
bigint operator*(const bigint& v) const {
vector<int> a4 = convert_base(this->z, base_digits, 4);
vector<int> b4 = convert_base(v.z, base_digits, 4);
vll a(a4.begin(), a4.end());
vll b(b4.begin(), b4.end());
while (a.size() < b.size()) a.push_back(0);
while (b.size() < a.size()) b.push_back(0);
while (a.size() & (a.size() - 1)) a.push_back(0), b.push_back(0);
vll c = karatsubaMultiply(a, b);
bigint res;
res.sign = sign * v.sign;
for (int i = 0, carry = 0; i < c.size(); i++) {
long long cur = c[i] + carry;
res.z.push_back((int)(cur % 10000));
carry = (int)(cur / 10000);
}
res.z = convert_base(res.z, 4, base_digits);
res.trim();
return res;
}
bigint& operator*=(const bigint& v) {
*this = *this * v;
return *this;
}
/*以上为高精度乘法所依赖项*/
bigint abs() const { return sign == 1 ? *this : -*this; }
friend pair<bigint, bigint> divmod(const bigint& a1, const bigint& b1) {
int norm = base / (b1.z.back() + 1);
bigint a = a1.abs() * norm;
bigint b = b1.abs() * norm;
bigint q, r;
q.z.resize(a.z.size());
for (int i = (int)a.z.size() - 1; i >= 0; i--) {
r *= base;
r += a.z[i];
int s1 = b.z.size() < r.z.size() ? r.z[b.z.size()] : 0;
int s2 = b.z.size() - 1 < r.z.size() ? r.z[b.z.size() - 1] : 0;
int d = (int)(((long long)s1 * base + s2) / b.z.back());
r -= b * d;
while (r < 0) r += b, --d;
q.z[i] = d;
}
q.sign = a1.sign * b1.sign;
r.sign = a1.sign;
q.trim();
r.trim();
return {q, r / norm};
}
bigint operator/(const bigint& v) const { return divmod(*this, v).first; }
bigint operator%(const bigint& v) const { return divmod(*this, v).second; }
bigint& operator/=(const bigint& v) {
*this = *this / v;
return *this;
}
/*以上为高精度除法所依赖项*/
bool operator>(const bigint& v) const { return v < *this; }
bool operator==(const bigint& v) const {
return !(*this < v) && !(v < *this);
}
/*以上为高精度快速幂所依赖项*/
bool isZero() const { return z.empty(); }
friend bigint gcd(const bigint& a, const bigint& b) {
return b.isZero() ? a : gcd(b, a % b);
}
friend bigint lcm(const bigint& a, const bigint& b) {
return a / gcd(a, b) * b;
}
/*以上为高精度gcd所依赖项*/
friend bigint sqrt(const bigint& a1) {
bigint a = a1;
while (a.z.empty() || a.z.size() % 2 == 1) a.z.push_back(0);
int n = (int)a.z.size();
int firstDigit = (int)::sqrt((double)a.z[n - 1] * base + a.z[n - 2]);
int norm = base / (firstDigit + 1);
a *= norm;
a *= norm;
while (a.z.empty() || a.z.size() % 2 == 1) a.z.push_back(0);
bigint r = (long long)a.z[n - 1] * base + a.z[n - 2];
firstDigit = (int)::sqrt((double)a.z[n - 1] * base + a.z[n - 2]);
int q = firstDigit;
bigint res;
for (int j = n / 2 - 1; j >= 0; j--) {
for (;; --q) {
bigint r1 =
(r - (res * 2 * base + q) * q) * base * base +
(j > 0 ? (long long)a.z[2 * j - 1] * base + a.z[2 * j - 2] : 0);
if (r1 >= 0) {
r = r1;
break;
}
}
res *= base;
res += q;
if (j > 0) {
int d1 = res.z.size() + 2 < r.z.size() ? r.z[res.z.size() + 2] : 0;
int d2 = res.z.size() + 1 < r.z.size() ? r.z[res.z.size() + 1] : 0;
int d3 = res.z.size() < r.z.size() ? r.z[res.z.size()] : 0;
q = (int)(((long long)d1 * base * base + (long long)d2 * base + d3) / (firstDigit * 2));
}
}
res.trim();
return res / norm;
}
};
bigint fpow (bigint a, bigint b) {
bigint res = 1;
while (b > 0) {
if (b % 2 == 1) res *= a;
a *= a;
b /= 2;
}
return res;
}
bigint a, b, c;
int main()
{
cin >> a >> b;
long long x = 1000000000000;
cout << a * b << endl;
cout << a - x << endl;
return 0;
}
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