高精度模板.
阿新 • • 發佈:2018-11-08
#include <iostream>
#include <string>
#include <cstring>
#include <cstdio>
using namespace std;
const int maxn = 1000;
struct bign{
int d[maxn], len;
void clean() { while(len > 1 && !d[len-1]) len--; }
bign() { memset(d, 0, sizeof (d)); len = 1; }
bign(int num) { *this = num; }
bign(char* num) { *this = num; }
bign operator = (const char* num){
memset(d, 0, sizeof(d)); len = strlen(num);
for(int i = 0; i < len; i++) d[i] = num[len-1-i] - '0';
clean();
return *this;
}
bign operator = (int num){
char s[20]; sprintf(s, "%d", num);
*this = s;
return *this;
}
bign operator + (const bign& b){
bign c = *this; int i;
for (i = 0; i < b.len; i++){
c.d[i] += b.d[i];
if (c.d[i] > 9) c.d[i]%=10 , c.d[i+1]++;
}
while (c.d[i] > 9) c.d[i++]%=10, c.d[i]++;
c.len = max(len, b.len);
if (c.d[i] && c.len <= i) c.len = i+1;
return c;
}
bign operator - (const bign& b){
bign c = *this; int i;
for (i = 0; i < b.len; i++){
c.d[i] -= b.d[i];
if (c.d[i] < 0) c.d[i]+=10, c.d[i+1]--;
}
while (c.d[i] < 0) c.d[i++]+=10, c.d[i]--;
c.clean();
return c;
}
bign operator * (const bign& b)const{
int i, j; bign c; c.len = len + b.len;
for(j = 0; j < b.len; j++)
for(i = 0; i < len; i++)
c.d[i+j] += d[i] * b.d[j];
for(i = 0; i < c.len-1; i++)
c.d[i+1] += c.d[i]/10, c.d[i] %= 10;
c.clean();
return c;
}
bign operator / (const bign& b){
int i, j;
bign c = *this, a = 0;
for (i = len - 1; i >= 0; i--)
{
a = a*10 + d[i];
for (j = 0; j < 10; j++) if (a < b*(j+1)) break;
c.d[i] = j;
a = a - b*j;
}
c.clean();
return c;
}
bign operator % (const bign& b){
int i, j;
bign a = 0;
for (i = len - 1; i >= 0; i--)
{
a = a*10 + d[i];
for (j = 0; j < 10; j++) if (a < b*(j+1)) break;
a = a - b*j;
}
return a;
}
bign operator += (const bign& b){
*this = *this + b;
return *this;
}
bool operator <(const bign& 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 false;
}
bool operator >(const bign& b) const{return b < *this;}
bool operator<=(const bign& b) const{return !(b < *this);}
bool operator>=(const bign& b) const{return !(*this < b);}
bool operator!=(const bign& b) const{return b < *this || *this < b;}
bool operator==(const bign& b) const{return !(b < *this) && !(b > *this);}
string str() const{
char s[maxn]={};
for(int i = 0; i < len; i++) s[len-1-i] = d[i]+'0';
return s;
}
};
istream& operator >> (istream& in, bign& x)
{
string s;
in >> s;
x = s.c_str();
return in;
}
ostream& operator << (ostream& out, const bign& x)
{
out << x.str();
return out;
}
int main()
{
freopen("noip.in","r",stdin);
freopen("noip.out","w",stdout);
bign s=0,t;
while (cin>>t)
{
if (t.len==1&&!t.d[0]) break;
s=s+t;
}
cout<<s<<endl;
return 0;
}
寫了個沒有/和後面運算的簡單一點的。。都是自己可以理解語法
/也挺簡單 但感覺noip用不到
#include <bits/stdc++.h> using namespace std; const int maxn=1000; #define IL inline struct bign{ int d[maxn],len; IL void clear() {while (len>1&&!d[len-1]) len--;} bign() {len=1;} bign(int num) { *this = num; } bign(char* num) { *this = num; } bign operator = (const char *num) { len=strlen(num); for (int i=0;i<len;i++) d[i]=num[len-i-1]-'0'; clear(); return *this; } bign operator =(int num) { char s[20]; sprintf(s,"%d",num); *this=s; return *this; } bign operator + (const bign b) { bign c=*this; int i; for (i=0;i<b.len;i++) c.d[i]+=b.d[i],c.d[i+1]+=c.d[i]/10,c.d[i]%=10; c.len=max(c.len,b.len); if (c.d[c.len]) c.len++; return c; } bign operator -(const bign b) { bign c=*this; int i; for (i=0;i<b.len;i++) { c.d[i]-=b.d[i]; if (c.d[i]<0) c.d[i]+=10,c.d[i+1]--; } while (c.d[i]<0) c.d[i]+=10,c.d[i+1]--,i++; return c; } bign operator *(const bign b) { int i,j; bign c; c.len=b.len+len; for (j=0;j<b.len;j++) for (i=0;i<len;i++) c.d[i+j]+=b.d[j]*d[i]; for (i=0;i<c.len;i++) c.d[i+1]+=c.d[i]/10,c.d[i]%=10; c.clear(); return c; } string str() { char s[maxn]={}; for (int i=0;i<len;i++) s[len-i-1]=d[i]+'0'; return s; } }; int main() { bign a=3424,b=213123,c=324,d=123,e=1344,f=132; bign ans=(a+b)*(c+d)*(e-f); cout<<ans.str(); return 0; }