大数运算和大数类
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1、首先看一个小问题：
整型数组、字符型数组、vector在未初始化的情况下：
int main(){int data[5];char c[5];vector<int> v(5);vector<char> v1(5);vector<int> v2;cout << "int[]:";for (int i = 0; i < 5; i++){cout << data[i] << " ";}cout << endl << endl;cout << "char[]:";for (int i = 0; i < 5; i++){cout << c[i] << ",";}cout << endl << endl;cout << "vector(int):";for (int i : v){cout << i << " ";}cout << endl << endl;cout << "vector(char):";for (auto i : v1){cout << i << ",";}cout << endl << endl;cout << "vector:";for (int i : v2){cout << i << ",";}cout << endl << endl;return 0;}
vs测试结果：
（1）整型数组在未初始化时，内存中是乱码；
（2）字符型数组在未初始化时，内存中是乱码；
（3）vector指定容量，int型时，初始化为0；
（4）vector指定容量，char型时，初始化为可打印的空白字符；
（5）vector不指定容量，默认0；
2、简单的大数相加：
#include<iostream>#include<string>#include<algorithm>using namespace std;#define N 500//25369874522455633 + 68452156965void bigNumAdd(string a, string b){int carry = 0; //表示进位//字符串反转(真正的数字位数 和 字符串下标 刚好是相反的，加法应该从数字低位算起) 且 字符串变数字int m[N] = { 0 };int n[N] = { 0 };int len1 = a.length(), len2 = b.length();for (int i = 0; i < len1; i++){m[i] = a[len1 - i - 1] - '0'; //加法应该从数字低位算起}for (int i = 0; i < len2; i++){n[i] = b[len2 - i - 1] - '0'; //加法应该从数字低位算起}//位运算————加法应该从数字低位算起int len = len1>len2 ? len1 + 1 : len2 + 1;string sum(len, 0);int k;for (k = 0; k < len1 || k < len2; k++){sum[k] = (m[k] + n[k] + carry) % 10 + '0';carry = (m[k] + n[k] + carry) / 10;}//处理最高位进位if (carry){sum[k] = carry + '0';}//结果字符串反转 并 输出reverse(sum.begin(), sum.end());cout << sum << endl;}int main(){string a, b;cin >> a >> b;bigNumAdd(a, b);return 0;}转自：http://blog.csdn.net/hackbuteer1/article/details/6595881
#include<iostream> #include<string> #include<iomanip> #include<algorithm> using namespace std; #define MAXN 9999 #define MAXSIZE 10 #define DLEN 4 class BigNum { private: int a[500]; //可以控制大数的位数 int len; //大数长度 public: BigNum(){ len = 1;memset(a,0,sizeof(a)); } //构造函数 BigNum(const int); //将一个int类型的变量转化为大数 BigNum(const char*); //将一个字符串类型的变量转化为大数 BigNum(const BigNum &); //拷贝构造函数 BigNum &operator=(const BigNum &); //重载赋值运算符，大数之间进行赋值运算 friend istream& operator>>(istream&, BigNum&); //重载输入运算符 friend ostream& operator<<(ostream&, BigNum&); //重载输出运算符 BigNum operator+(const BigNum &) const; //重载加法运算符，两个大数之间的相加运算 BigNum operator-(const BigNum &) const; //重载减法运算符，两个大数之间的相减运算 BigNum operator*(const BigNum &) const; //重载乘法运算符，两个大数之间的相乘运算 BigNum operator/(const int &) const; //重载除法运算符，大数对一个整数进行相除运算 BigNum operator^(const int &) const; //大数的n次方运算 int operator%(const int &) const; //大数对一个int类型的变量进行取模运算 bool operator>(const BigNum & T)const; //大数和另一个大数的大小比较 bool operator>(const int & t)const; //大数和一个int类型的变量的大小比较 void print(); //输出大数 }; BigNum::BigNum(const int b) //将一个int类型的变量转化为大数 { int c,d = b; len = 0; memset(a,0,sizeof(a)); while(d > MAXN) { c = d - (d / (MAXN + 1)) * (MAXN + 1); d = d / (MAXN + 1); a[len++] = c; } a[len++] = d; } BigNum::BigNum(const char*s) //将一个字符串类型的变量转化为大数 { int t,k,index,l,i; memset(a,0,sizeof(a)); l=strlen(s); len=l/DLEN; if(l%DLEN) len++; index=0; for(i=l-1;i>=0;i-=DLEN) { t=0; k=i-DLEN+1; if(k<0) k=0; for(int j=k;j<=i;j++) t=t*10+s[j]-'0'; a[index++]=t; } } BigNum::BigNum(const BigNum & T) : len(T.len) //拷贝构造函数 { int i; memset(a,0,sizeof(a)); for(i = 0 ; i < len ; i++) a[i] = T.a[i]; } BigNum & BigNum::operator=(const BigNum & n) //重载赋值运算符，大数之间进行赋值运算 { int i; len = n.len; memset(a,0,sizeof(a)); for(i = 0 ; i < len ; i++) a[i] = n.a[i]; return *this; } istream& operator>>(istream & in, BigNum & b) //重载输入运算符 { char ch[MAXSIZE*4]; int i = -1; in>>ch; int l=strlen(ch); int count=0,sum=0; for(i=l-1;i>=0;) { sum = 0; int t=1; for(int j=0;j<4&&i>=0;j++,i--,t*=10) { sum+=(ch[i]-'0')*t; } b.a[count]=sum; count++; } b.len =count++; return in; } ostream& operator<<(ostream& out, BigNum& b) //重载输出运算符 { int i; cout << b.a[b.len - 1]; for(i = b.len - 2 ; i >= 0 ; i--) { cout.width(DLEN); cout.fill('0'); cout << b.a[i]; } return out; } BigNum BigNum::operator+(const BigNum & T) const //两个大数之间的相加运算 { BigNum t(*this); int i,big; //位数 big = T.len > len ? T.len : len; for(i = 0 ; i < big ; i++) { t.a[i] +=T.a[i]; if(t.a[i] > MAXN) { t.a[i + 1]++; t.a[i] -=MAXN+1; } } if(t.a[big] != 0) t.len = big + 1; else t.len = big; return t; } BigNum BigNum::operator-(const BigNum & T) const //两个大数之间的相减运算 { int i,j,big; bool flag; BigNum t1,t2; if(*this>T) { t1=*this; t2=T; flag=0; } else { t1=T; t2=*this; flag=1; } big=t1.len; for(i = 0 ; i < big ; i++) { if(t1.a[i] < t2.a[i]) { j = i + 1; while(t1.a[j] == 0) j++; t1.a[j--]--; while(j > i) t1.a[j--] += MAXN; t1.a[i] += MAXN + 1 - t2.a[i]; } else t1.a[i] -= t2.a[i]; } t1.len = big; while(t1.a[t1.len - 1] == 0 && t1.len > 1) { t1.len--; big--; } if(flag) t1.a[big-1]=0-t1.a[big-1]; return t1; } BigNum BigNum::operator*(const BigNum & T) const //两个大数之间的相乘运算 { BigNum ret; int i,j,up; int temp,temp1; for(i = 0 ; i < len ; i++) { up = 0; for(j = 0 ; j < T.len ; j++) { temp = a[i] * T.a[j] + ret.a[i + j] + up; if(temp > MAXN) { temp1 = temp - temp / (MAXN + 1) * (MAXN + 1); up = temp / (MAXN + 1); ret.a[i + j] = temp1; } else { up = 0; ret.a[i + j] = temp; } } if(up != 0) ret.a[i + j] = up; } ret.len = i + j; while(ret.a[ret.len - 1] == 0 && ret.len > 1) ret.len--; return ret; } BigNum BigNum::operator/(const int & b) const //大数对一个整数进行相除运算 { BigNum ret; int i,down = 0; for(i = len - 1 ; i >= 0 ; i--) { ret.a[i] = (a[i] + down * (MAXN + 1)) / b; down = a[i] + down * (MAXN + 1) - ret.a[i] * b; } ret.len = len; while(ret.a[ret.len - 1] == 0 && ret.len > 1) ret.len--; return ret; } int BigNum::operator %(const int & b) const //大数对一个int类型的变量进行取模运算 { int i,d=0; for (i = len-1; i>=0; i--) { d = ((d * (MAXN+1))% b + a[i])% b; } return d; } BigNum BigNum::operator^(const int & n) const //大数的n次方运算 { BigNum t,ret(1); int i; if(n<0) exit(-1); if(n==0) return 1; if(n==1) return *this; int m=n; while(m>1) { t=*this; for( i=1;i<<1<=m;i<<=1) { t=t*t; } m-=i; ret=ret*t; if(m==1) ret=ret*(*this); } return ret; } bool BigNum::operator>(const BigNum & T) const //大数和另一个大数的大小比较 { int ln; if(len > T.len) return true; else if(len == T.len) { ln = len - 1; while(a[ln] == T.a[ln] && ln >= 0) ln--; if(ln >= 0 && a[ln] > T.a[ln]) return true; else return false; } else return false; } bool BigNum::operator >(const int & t) const //大数和一个int类型的变量的大小比较 { BigNum b(t); return *this>b; } void BigNum::print() //输出大数 { int i; cout << a[len - 1]; for(i = len - 2 ; i >= 0 ; i--) { cout.width(DLEN); cout.fill('0'); cout << a[i]; } cout << endl; } int main(void) { int i,n; BigNum x[101]; //定义大数的对象数组 x[0]=1; for(i=1;i<101;i++) x[i]=x[i-1]*(4*i-2)/(i+1); while(scanf("%d",&n)==1 && n!=-1) { x[n].print(); } } 附录：
#include <iostream>#include <string>using namespace std;int compare(string str1, string str2){//相等返回0，大于返回1，小于返回-1if (str1.size()>str2.size()) return 1; //长度长的整数大于长度小的整数else if (str1.size()<str2.size()) return -1;else return str1.compare(str2); //若长度相等，则头到尾按位比较}string SUB_INT(string str1, string str2);string ADD_INT(string str1, string str2) {//高精度加法int sign = 1; //sign 为符号位string str;if (str1[0] == '-') {if (str2[0] == '-') {sign = -1;str = ADD_INT(str1.erase(0, 1), str2.erase(0, 1));}else {str = SUB_INT(str2, str1.erase(0, 1));}}else {if (str2[0] == '-') {str = SUB_INT(str1, str2.erase(0, 1));}else { //把两个整数对齐，短整数前面加0补齐string::size_type L1, L2;int i;L1 = str1.size();L2 = str2.size();if (L1<L2) {for (i = 1; i <= L2 - L1; i++) str1 = "0" + str1;}else {for (i = 1; i <= L1 - L2; i++) str2 = "0" + str2;}int int1 = 0, int2 = 0; //int2 记录进位for (i = str1.size() - 1; i >= 0; i--) {int1 = (int(str1[i]) - '0' + int(str2[i]) - '0' + int2) % 10;int2 = (int(str1[i]) - '0' + int(str2[i]) - '0' + int2) / 10;str = char(int1 + '0') + str;}if (int2 != 0) str = char(int2 + '0') + str;}}//运算后处理符号位if ((sign == -1) && (str[0] != '0')) str = "-" + str;return str;}string SUB_INT(string str1, string str2) {//高精度减法int sign = 1; //sign 为符号位string str;int i, j;if (str2[0] == '-') {str = ADD_INT(str1, str2.erase(0, 1));}else {int res = compare(str1, str2);if (res == 0) return "0";if (res<0) {sign = -1;string temp = str1;str1 = str2;str2 = temp;}string::size_type tempint;tempint = str1.size() - str2.size();for (i = str2.size() - 1; i >= 0; i--) {if (str1[i + tempint]<str2[i]) {j = 1;while (1) {//zhao4zhong1添加if (str1[i + tempint - j] == '0') {str1[i + tempint - j] = '9';j++;}else {str1[i + tempint - j] = char(int(str1[i + tempint - j]) - 1);break;}}str = char(str1[i + tempint] - str2[i] + ':') + str;}else {str = char(str1[i + tempint] - str2[i] + '0') + str;}}for (i = tempint - 1; i >= 0; i--) str = str1[i] + str;}//去除结果中多余的前导0str.erase(0, str.find_first_not_of('0'));if (str.empty()) str = "0";if ((sign == -1) && (str[0] != '0')) str = "-" + str;return str;}string MUL_INT(string str1, string str2) { //高精度乘法int sign = 1; //sign 为符号位string str;if (str1[0] == '-') {sign *= -1;str1 = str1.erase(0, 1);}if (str2[0] == '-') {sign *= -1;str2 = str2.erase(0, 1);}int i, j;string::size_type L1, L2;L1 = str1.size();L2 = str2.size();for (i = L2 - 1; i >= 0; i--){//模拟手工乘法竖式string tempstr;int int1 = 0, int2 = 0, int3 = int(str2[i]) - '0';if (int3 != 0) {for (j = 1; j <= (int)(L2 - 1 - i); j++)tempstr = "0" + tempstr;for (j = L1 - 1; j >= 0; j--){int1 = (int3*(int(str1[j]) - '0') + int2) % 10;int2 = (int3*(int(str1[j]) - '0') + int2) / 10;tempstr = char(int1 + '0') + tempstr;}if (int2 != 0) tempstr = char(int2 + '0') + tempstr;}str = ADD_INT(str, tempstr);}//去除结果中的前导0str.erase(0, str.find_first_not_of('0'));if (str.empty()) str = "0";if ((sign == -1) && (str[0] != '0'))str = "-" + str;return str;}string DIVIDE_INT(string str1, string str2, int flag) { //高精度除法。flag==1时,返回商; flag==0时,返回余数string quotient, residue; //定义商和余数int sign1 = 1, sign2 = 1;if (str2 == "0") { //判断除数是否为0quotient = "ERROR!";residue = "ERROR!";if (flag == 1) return quotient;else return residue;}if (str1 == "0") { //判断被除数是否为0quotient = "0";residue = "0";}if (str1[0] == '-') {str1 = str1.erase(0, 1);sign1 *= -1;sign2 = -1;}if (str2[0] == '-') {str2 = str2.erase(0, 1);sign1 *= -1;}int res = compare(str1, str2);if (res<0) {quotient = "0";residue = str1;}else if (res == 0) {quotient = "1";residue = "0";}else {string::size_type L1, L2;L1 = str1.size();L2 = str2.size();string tempstr;tempstr.append(str1, 0, L2 - 1);for (int i = L2 - 1; i<L1; i++) { //模拟手工除法竖式tempstr = tempstr + str1[i];tempstr.erase(0, tempstr.find_first_not_of('0'));//zhao4zhong1添加if (tempstr.empty()) tempstr = "0";//zhao4zhong1添加for (char ch = '9'; ch >= '0'; ch--) { //试商string str;str = str + ch;if (compare(MUL_INT(str2, str), tempstr) <= 0) {quotient = quotient + ch;tempstr = SUB_INT(tempstr, MUL_INT(str2, str));break;}}}residue = tempstr;}//去除结果中的前导0quotient.erase(0, quotient.find_first_not_of('0'));if (quotient.empty()) quotient = "0";if ((sign1 == -1) && (quotient[0] != '0')) quotient = "-" + quotient;if ((sign2 == -1) && (residue[0] != '0')) residue = "-" + residue;if (flag == 1) return quotient;else return residue;}string DIV_INT(string str1, string str2) {//高精度除法,返回商return DIVIDE_INT(str1, str2, 1);}string MOD_INT(string str1, string str2) {//高精度除法,返回余数return DIVIDE_INT(str1, str2, 0);}int main(){/*char ch;string s1, s2, res;while (cin >> s1 >> ch >> s2) {switch (ch) {case '+':res = ADD_INT(s1, s2); break;case '-':res = SUB_INT(s1, s2); break;case '*':res = MUL_INT(s1, s2); break;case '/':res = DIV_INT(s1, s2); break;case '%':res = MOD_INT(s1, s2); break;default: break;}cout << res << endl;}return(0);*/string s11, s22, res1;s11 = "1";s22 = "2";for (int i = 1; i <= 1000; i++){res1 = MUL_INT(s11, s22); //利用乘法做幂次运算！s11 = res1;}cout << res1 << endl;return(0);} 0 0
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