C++的大数除法最快速度的算法PS：用顺序表做个容器将每位数都放进去,对其进行大数的运算,用的是循环

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通过移位、取模来大大提高效率,只是程序比较难理解,
需要慢慢研究
#include
#include
#include
using namespace std;
const int base = 10000; // (base^2) fit into int
const int width = 4; // width = log base
const int N = 1000; // n * width:可表示的最大位数
struct bint{
int ln,v[N];
bint (int r = 0) { // r应该是字符串!
for (ln = 0; r > 0; r /= base) v[ln++] = r % base;
}bint& operator = (const bint& r) {
memcpy(this,&r,(r.ln + 1) * sizeof(int));// !
return *this;
}};bool operator int i;if (a.ln != b.ln) return a.ln for (i = a.ln - 1; i >= 0 && a.v[i] == b.v[i]; i--);
return i }bool operator 0; i++) {
if (i if (i res.v[i] = cy % base; cy /= base;
}res.ln = i;
return res;
}bint operator - (const bint& a,const bint& b){
bint res; int i,cy = 0;
for (res.ln = a.ln,i = 0; i res.v[i] = a.v[i] - cy;
if (i if (res.v[i] else cy = 0;
}while (res.ln > 0 && res.v[res.ln - 1] == 0) res.ln--;return res;
}bint operator * (const bint& a,const bint& b){bint res; res.ln = 0;if (0 == b.ln) { res.v[0] = 0; return res; }int i,j,cy;for (i = 0; i for (j=cy=0; j 0; j++,cy/= base) {if (j if (i + j if (i + j >= res.ln) res.v[res.ln++] = cy % base;else res.v[i + j] = cy % base;}}return res;
}bint operator / (const bint& a,const bint& b){ // !b != 0bint tmp,mod,res;int i,lf,rg,mid;mod.v[0] = mod.ln = 0;