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

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

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 intconst int width = 4; // width = log baseconst 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;

这下我知道了