《程序员密码学》之大数算数&Eratosthenes素数筛选
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10进制大数算法，支持+、-、*、/、%以及一般的比较运算符，支持字符串方式读入以及输出大数
#pragma once#include <iostream>#include <string>using namespace std;#define SWAP(x,y,t) ((t)=(x),(x)=(y),(y)=(x))#define MIN(x,y) (x)>(y)?(y):(x)#define MAX(x,y) (x)>(y)?(x):(y)#define BASE 10000 //大数以每位十进制数的方式存储//乘法实现，计算A*B，结果返回到 C，函数结果为乘法结果的长度#define MAX_SIZE 128//一般将算术和关系操作符定义非成员函数，而将赋值操作符定义为成员：//大数以小端格式存储struct BigNum{int len; int v[MAX_SIZE];BigNum(){ len=0;memset(v,0, sizeof(v));}BigNum(const char* num){*this=num;}BigNum(int num){len=0;for(; num>0; num/=BASE) v[len++] = num%BASE;}BigNum&operator=(const char* num){intl=strlen(num);intn=(l%4)?(l%4):4;len=0;charx[5]={0};memcpy(x,num, n );v[(l+4-1)/4-1]= atoi(x);len++;for(int i=(l+4-1)/4-2,j=n; i>=0; i--,j+=4){memcpy(x, num+j, 4);v[i]= atoi(x);len++;}return *this;}};booloperator<(const BigNum& left,const BigNum& right){int i;if(left.len!=right.len)return left.len<right.len;for(i=left.len-1; i>=0 && left.v[i]==right.v[i]; i--);return i>0? left.v[i]<right.v[i]: 0;}booloperator>(const BigNum& left,const BigNum& right){return right<left;}booloperator==(const BigNum& left,const BigNum& right){return (!(left<right)) && (!(right<left));}booloperator<=(const BigNum& left,const BigNum& right){return (left<right) || (left==right);}booloperator>=(const BigNum& left,const BigNum& right){return right<=left;}BigNumoperator+(const BigNum& left,const BigNum& right){int n = MAX(left.len,right.len);int x=0,i;BigNumresult;for(i=0; i<n; i++){if(i<left.len) x+=left.v[i];if(i<right.len) x+=right.v[i];result.v[i] = x%BASE;x/=BASE;}if(x!=0)result.v[i++] = x;result.len= i;return result;}BigNumoperator-(const BigNum& left,const BigNum& right){BigNumresult;int i,x=0;for(i=0; i<left.len; i++){result.v[i] = left.v[i] - x;if(i<right.len) result.v[i] -= right.v[i];if(result.v[i]<0)result.v[i] += BASE, x = 1;elsex=0;}result.len= i;while(result.len>0&&result.v[result.len-1]==0)result.len--;return result;}BigNum operator*(const BigNum& left,const BigNum& right){ BigNum result; int c0,c1,c2;//存数临时结果 int tx,ty; int len=left.len+right.len; int *templ,*tempr; c0 = c1 = c2 = 0; for(int i=0; i<len; i++)//i为本次计算在结果中所处的级别，这里一个整形为一个级别 { ty = MIN(i, right.len-1); tx = i-ty; templ = (int*)(left.v + tx); tempr = (int*)(right.v+ ty); int n = MIN( left.len-tx, ty+1);//本级别可计算的次数 c0 = c1; c1 = c2; c2 = 0; //上次计算结果向下移一个级别，进入本次计算 for(int j=0; j<n; j++) { int x=*templ++, y=*tempr--; c0 += x*y; /*_asm //mul operation { pushad mov eax,x mov ebx,y mul ebx //无符号乘法指令 add c0,eax //adc c1,edx //adc c2,0 popad }*/ } result.v[i]=c0%BASE; c1+=c0/BASE; c2+=c1/BASE; } //计算长度 for(int i=len-1; i >= 0; i--) if (result.v[i]==0) len--; else break; result.len = len; return result;}BigNumoperator/(const BigNum& a,const BigNum& b){BigNum tmp, mod, res;int i, lf, rg, mid,base=BASE;mod.v[0] = mod.len = 0;for (i = a.len - 1; i >= 0; i--) {//从最高4位开始，每次增加一个4位试商mod = mod * base + a.v[i];//二分法寻找mod/b的商for (lf = 0, rg = base -1; lf < rg; ) {mid = (lf + rg + 1) / 2;if (b * mid <= mod) lf = mid;else rg = mid - 1;}//存储结果并将剩余的数继续试商res.v[i] = lf;mod = mod - b * lf;}res.len = a.len;while (res.len > 0 && res.v[res.len - 1] == 0) res.len--;return res; // return mod 就是%运算}BigNumoperator%(const BigNum& a,const BigNum& b){BigNum tmp, mod, res;int i, lf, rg, mid,base=BASE;mod.v[0] = mod.len = 0;for (i = a.len - 1; i >= 0; i--) {//从最高4位开始，每次增加一个4位试商mod = mod * base + a.v[i];//二分法寻找mod/b的商for (lf = 0, rg = base -1; lf < rg; ) {mid = (lf + rg + 1) / 2;if (b * mid <= mod) lf = mid;else rg = mid - 1;}//存储结果并将剩余的数继续试商res.v[i] = lf;mod = mod - b * lf;}res.len = a.len;while (res.len > 0 && res.v[res.len - 1] == 0) res.len--;return mod;// %运算}ostream& operator<<(ostream& os,const BigNum& num){ printf( "%d",num.v[num.len-1]); for(int i=num.len-2; i>=0; i--) printf( "%04d",num.v[i]); return os;}istream& operator>>(istream& in,BigNum& num){string str;in>>str;num=str.c_str();return in;}X86版本的大数乘法简陋实现，没有书上的可移植性,结果以16进制存储
只是为了试试效果~
#include <stdio.h>#define MIN(x,y) (x)>(y)?(y):(x)/*#define MUL(x,y) _asm {\ "pushad\r\n"\ "mov eax,x\r\n" \ "mov ebx,y\r\n" \ "mul ebx\r\n"\ "add c0,eax\r\n" \ "adc c1,edx\r\n" \ "adc c2,0\r\n"\ "popad\r\n"\ }*///乘法实现，计算A*B，结果返回到C，函数结果为乘法结果的长度，大数整体采用小端模式存储int mul(int *A,int lena, int *B,int lenb, int *C){ int ix, iy, iz, tx, ty, pa; int c0, c1, c2, *tmpx, *tmpy; c0=c1=c2=0; /* get size of output and trim */ pa = lena+lenb; for (ix = 0; ix < pa; ix++) { /* get offsets into the two bignums */ ty = MIN(ix, lenb-1); tx = ix - ty; /* setup temp aliases */ tmpx = A + tx; tmpy = B + ty; /* this is the number of times the loop will iterrate, essentially its while (tx++ < a->used && ty-- >= 0) { ... } */ iy = MIN(lena-tx, ty+1); /* execute loop */ c0=c1;c1=c2;c2=0; for (iz = 0; iz < iy; ++iz) { int i=*tmpx++,j= *tmpy--; _asm //mul operation { pushad mov eax,i mov ebx,j mul ebx add c0,eax adc c1,edx adc c2,0 popad } } /* store term */ C[ix]=c0; } return pa;}//平方计算，函数结果为乘法结果的长度int square(int *A,int lena, int *C){ int ix, iy, iz, tx, ty, pa; int c0, c1, c2, *tmpx, *tmpy; c0=c1=c2=0; /* get size of output and trim */ pa = lena*2; for (ix = 0; ix < pa; ix++) { /* get offsets into the two bignums */ ty = MIN(ix, lena-1); tx = ix - ty; /* setup temp aliases */ tmpx = A + tx; tmpy = A + ty; /* this is the number of times the loop will iterrate, essentially its while (tx++ < a->used && ty-- >= 0) { ... } */ iy = MIN(lena-tx, ty+1); iy = MIN(iy, (ty-tx+1)>>1); /* execute loop */ c0=c1;c1=c2;c2=0; for (iz = 0; iz < iy; ++iz) { int i=*tmpx++,j= *tmpy--; _asm // 由于是平方，加两次，省去一半的乘法指令 { pushad mov eax,i mov ebx,j mul ebx add c0,eax adc c1,edx adc c2,0 add c0,eax adc c1,edx adc c2,0 popad } } if ((ix&1)==0) // 偶数位的话需要加上中间未重复的结果 { int i=A[ix>>1],j= A[ix>>1]; _asm //mul operation { pushad mov eax,i mov ebx,j mul ebx add c0,eax adc c1,edx adc c2,0 popad } } /* store term */ C[ix]=c0; } return pa;}//求幂,lena:A的长度(4字节为单位),exponent:指数,C:存储结果//这个版本的求幂只能求2^n形式的幂int power(int *A,int lena, int exponent, int *C){ int len=lena,*temp,*dst=C; int squ=0;// 可直接平方的次数 for(int t=exponent; t!=1; t>>=1) squ++; for(int i=0; i < squ; i++) { memset(C, 0 , 1024); len = square(A,len, C); SWAP(A,C,temp); } if (dst!=A) memcpy(dst,A,len); return len;}void print_bignum(int *bignum,int u){ int first=1; printf( "0x"); for(int i=u-1; i >= 0; i--) if ((bignum[i]==0)&&first) continue; else { first = 0; printf( "%08X",bignum[i]); }}int main(){ int a[] = {0x87654321,0x12345678,0x12345678,0x12345678,0x12345678,0x12345678,0x12345678,\ 0x87654321,0x12345678,0x12345678,0x12345678,0x12345678,0x12345678,0x12345678,\ 0x12345678,0x12345678,0x12345678,0x12345678,0x12345678,0x12345678,0x12345678,0x12345678,0x12345678,0x12345678,0x12345678}; int b[] = {0x87654321,0x12345678,0x12345678,0x12345678,0x12345678,0x12345678,\ 0x87654321,0x12345678,0x12345678,0x12345678,0x12345678,0x12345678,0x12345678,\ 0x12345678,0x12345678,0x12345678,0x12345678,0x12345678,0x12345678,0x12345678,0x12345678,0x12345678,0x12345678,0x12345678,0x12345678}; int c[256]={0}; int len=mul(a,sizeof (a)/sizeof( int),b,sizeof (b)/sizeof( int), c); print_bignum(c,len); getchar(); return 0;}
#include <iostream>#include <cmath>#include <vector>using namespace std;//查找指定范围内的所有素数//100以内的所有素数unsigned int prime_tab[] = { 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97};//Eratosthenes 素数筛选,取得0-num内所有的素数,c++容器版int Eratosthenes(vector<int >& tab, unsigned int num){#define TEST(p,x) (((p)%(x))!=0) //是否满足素数条件 if(num <= 100) { for(int i=0; i < sizeof(prime_tab)/ sizeof(int ); i++) { if(prime_tab[i]<num) tab.push_back(prime_tab[i]); else break; } return tab.size(); } unsigned int i=(int)sqrt(( double)num); Eratosthenes(tab, i); /*递归初始化i以内的素数表 */ for(; i < num; i++)// 筛选 { bool flag=false ; int n=tab.size(); for(vector<int >::iterator iter=tab.begin(),end=iter+n; iter != end ; iter++) { if( !TEST(i,*iter) ) break; if( (iter+1)==end ) flag= true; } if(flag) tab.push_back(i); }#undef TEST(p,x) return tab.size();}//Eratosthenes 素数筛选,取得 num内所有的素数int Eratosthenes2(unsigned int num){ unsigned int count=0; char *tab = new char[num+1]; memset(tab+2,1, num-2); for(unsigned int i=2; i <= ( unsigned int )sqrt((double)num); i++) if(tab[i]==1) for(unsigned int j=i; j*i <= num; j++) tab[i*j] = 0; for(unsigned int i=2; i <= num ; i++) if(tab[i]==1) { printf( "%d\t",i); count++; } printf( "\n%d ",count); delete [] tab; return count;}int main(){ vector< int> tab; //素数表容器 tab.reserve(2000); //预留空间 cout<<Eratosthenes(tab,150000); //函数返回统计个数 for(int i=0; i < tab.size(); i++) //输出素数 printf( "%d\t", tab[i]); getchar(); return 0;}
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