高精度模板
Posted Jozky86
tags:
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看到一个非常不错的高精度模板，记录一下
#include<cstdio>
#include<cstring>
using namespace std;
typedef long long ll;
const int base = 1e8;
const int N = 1e4 + 10;
int aux[N << 3];
struct bigint
int s[N], l;
void CL() l = 0; memset(s, 0, sizeof(s));
void pr()
printf("%d", s[l]);
for (int i = l - 1; i; i--)
printf("%08d", s[i]);
void re_l()
int i, x = 0, k = 1, L = 0, fl, o;
char c = getchar();
for (; c < '0' || c > '9'; c = getchar());
for (; c >= '0' && c <= '9'; c = getchar())
if (!(L - 1) && !aux[L])
L--;
aux[++L] = c - '0';
CL();
l = L / 8 + ((o = L % 8) > 0);
for (i = 1; i <= o; i++)
x = x * 10 + aux[i];
if (o)
s[l] = x;
for (fl = !o ? l + 1 : l, i = o + 1, x = 0; i <= L; i++, k++)
x = x * 10 + aux[i];
if (!(k ^ 8))
s[--fl] = x, x = k = 0;
if (!l)
l = 1;
ll toint()
ll x = 0;
for (int i = l; i; i--)
x = x * base + s[i];
return x;
bigint operator = (int b)
CL();
do
s[++l] = b % base;
b /= base;
while (b > 0);
return *this;
bigint operator = (ll b)
CL();
do
s[++l] = b % base;
b /= base;
while (b > 0);
return *this;
bigint operator + (const int& b)
bigint c = *this;
ll x = b;
for (int i = 1; i <= l && x; i++)
x = x + c.s[i];
c.s[i] = x % base;
x /= base;
if (x)
c.s[++c.l] = x;
return c;
bigint operator + (const ll & b)
bigint c = *this;
ll x = b;
for (int i = 1; i <= l && x; i++)
x = x + c.s[i];
c.s[i] = x % base;
x /= base;
if (x)
c.s[++c.l] = x;
return c;
bigint operator + (bigint & b)
if (b.l < 3)
return *this + b.toint();
bigint c;
ll x = 0;
int k = l < b.l ? b.l : l;
c.CL(); c.l = k;
for (int i = 1; i <= k; i++)
x = x + s[i] + b.s[i];
c.s[i] = x % base;
x /= base;
if (x)
c.s[++c.l] = x;
return c;
bigint operator - (const bigint & b)
bigint c, d = *this;
ll x = 0;
c.CL();
for (int i = 1; i <= l; i++)
if ((x = d.s[i]) < b.s[i])
d.s[i + 1]--;
x += base;
c.s[i] = x - b.s[i];
c.l = l;
for (; !c.s[c.l] && c.l > 1; c.l--);
return c;
bigint operator - (const int& b) bigint c; return *this - (c = b);
bigint operator - (const ll & b) bigint c; return *this - (c = b);
bigint operator * (const int& b)
bigint c;
ll x = 0;
c.CL();
for (int i = 1; i <= l; i++)
x = x + 1LL * s[i] * b;
c.s[i] = x % base;
x /= base;
for (c.l = l; x; x /= base)
c.s[++c.l] = x % base;
return c;
bigint operator * (bigint & b)
if (b.l < 2)
return *this * b.toint();
bigint c;
ll x;
int i, j, k;
c.CL();
for (i = 1; i <= l; i++)
x = 0;
for (j = 1; j <= b.l; j++)
x = x + 1LL * s[i] * b.s[j] + c.s[k = i + j - 1];
c.s[k] = x % base;
x /= base;
if (x)
c.s[i + b.l] = x;
for (c.l = l + b.l; !c.s[c.l] && c.l > 1; c.l--);
return c;
bigint operator * (const ll & b)
bigint c;
if (b > 2e9)
c = b;
return *this* c;
ll x = 0;
c.CL();
for (int i = 1; i <= l; i++)
x = x + b * s[i];
c.s[i] = x % base;
x /= base;
for (c.l = l; x; x /= base)
c.s[++c.l] = x % base;
return c;
bigint operator / (const int& b)
bigint c;
ll x = 0;
c.CL();
for (int i = l; i; i--)
c.s[i] = (x * base + s[i]) / b;
x = (x * base + s[i]) % b;
for (c.l = l; !c.s[c.l] && c.l > 1; c.l--);
return c;
bigint operator / (const ll & b)
bigint c;
ll x = 0;
c.CL();
for (int i = l; i; i--)
c.s[i] = (x * base + s[i]) / b;
x = (x * base + s[i]) % b;
for (c.l = l; !c.s[c.l] && c.l > 1; c.l--);
return c;
bigint operator / (bigint & b)
if (b.l < 2)
return *this / b.toint();
bigint c, d;
int i, j, le, r, mid, k;
c.CL(); d.CL();
for (i = l; i; i--)
for (j = ++d.l; j > 1; j--)
d.s[j] = d.s[j - 1];
d.s[1] = s[i];
if (d < b)
continue;
le = k = 0; r = base - 1;
while (le <= r)
mid = (le + r) >> 1;
b * mid <= d ? le = mid + 1, k = mid : r = mid - 1;
c.s[i] = k; d = d - b * k;
for (c.l = l; !c.s[c.l] && c.l > 1; c.l--);
return c;
bigint operator % (const int& b)
bigint c;
ll x = 0;
c.CL();
for (int i = l; i; i--)
x = (x * base + s[i]) % b;
return c = x;
bigint operator % (const ll & b)
bigint c;
ll x = 0;
c.CL();
for (int i = l; i; i--)
x = (x * base + s[i]) % b;
return c = x;
bigint operator % (bigint & b)
if (b.l < 2)
return *this % b.toint();
bigint c;
int i, j, le, r, mid, k;
c.CL();
for (i = l; i; i--)
for (j = ++c.l; j > 1; j--)
c.s[j] = c.s[j - 1];
c.s[1] = s[i];
if (c < b)
continue;
le = k = 0; r = base - 1;
while (le <= r)
mid = (le + r) >> 1;
b * mid <= c ? le = mid + 1, k = mid : r = mid - 1;
c = c - b * k;
for (; !c.s[c.l] && c.l > 1; c.l--);
return c;
bigint operator += (bigint & b) return *this = *this + b;
bigint operator += (ll b) return *this = *this + b;
bigint operator += (int b) return *this = *this + b;
bigint operator -= (bigint & b) return *this = *this - b;
bigint operator -= (ll b) return *this = *this - b;
bigint operator -= (int b) return *this = *this - b;
bigint operator *= (bigint & b) return *this = *this * b;
bigint operator *= (ll b) return *this = *this * b;
bigint operator *= (int b) return *this = *this * b;
bigint operator /=以上是关于高精度模板的主要内容，如果未能解决你的问题，请参考以下文章