2066 - A+B again
Your job is very easy !!!
Just Calculate A + B.
Input
Each line will contain two integers A and B. Process to end of file.And both A and B you can use "long long"to store them
Output
For each case, output A + B in one line.
Examples
Input
1 1 10 10 11111111111 11111111111
Output
2 20 22222222222
Hint
Bazinga !!!
Solution C
//求两个整数想加的结果，这两个整数都可以用long long 存下，但是它们的结果不一定可以用 //long long 存下，所以要用大整数来求解 #include<stdio.h> #include<string.h> int main() { char strA[50], strB[50], strC[50]; int C[51], lenA, lenB; while(scanf("%s%s", strA, strB) != EOF){ memset(C, 0, sizeof(C)); //这一段完成的是A和B都为正数的时候的情况 if((strA[0]>='0' && strA[0] <= '9') && (strB[0]>='0' && strB[0] <= '9')){ lenA = strlen(strA); lenB = strlen(strB); int start = 0; int ch = 0; //ch是用来标记有没有进位的 int i, j; for(i = lenA - 1,j = lenB - 1; i >= 0 && j >= 0; i--, j--) { C[start] = ch + strA[i] + strB[j] - '0' - '0'; if(C[start] > 9) {C[start] = C[start] - 10; ch = 1;} else ch = 0; start++; if(i == 0 && j == 0 && ch == 1) C[start++] = 1; //这里需要注意是因为如果A[],B[]同时到了第一个数 } //并且此时产生了进位的话，那就需要在数组C[]中表现出来 if(i > 0) while(i >= 0) {C[start++] = ch + strA[i--] - '0'; ch = 0;} if(j > 0) while(j >= 0) {C[start++] = ch + strB[j--] - '0'; ch = 0;} for(int i = start-1; i >= 0; i--) printf("%d", C[i]); //把最后的结果输出 printf("\n"); }//if //当A和B都为负数的情况，可以转化为求正数的东西来解 else if(strA[0] == '-' && strB[0] == '-'){ lenA = strlen(strA); lenB = strlen(strB); int start = 0; int ch = 0; int i, j; for(i = lenA - 1,j = lenB - 1; i >= 1 && j >= 1; i--, j--) { C[start] = ch + strA[i] + strB[j] - '0' - '0'; if(C[start] > 9) {C[start] = C[start] - 10; ch = 1;} else ch = 0; start++; if(i == 1 && j == 1 && ch == 1) C[start++] = 1; } if(i > 1) while(i >= 1) {C[start++] = ch + strA[i--] - '0'; ch = 0;} if(j > 1) while(j >= 1) {C[start++] = ch + strB[j--] - '0'; ch = 0;} printf("-"); for(int i = start-1; i >= 0; i--) printf("%d", C[i]); printf("\n"); }//else if else{//如果A和B不同时为正数或者同时为负数的话就把它们转化为A为正数B为负数的情况 if((strB[0]>='0' && strB[0] <= '9') && strA[0] == '-'){ memset(strC, 0, sizeof(strC)); strncpy(strC, strA, sizeof(strA)); strncpy(strA, strB, sizeof(strB)); strncpy(strB, strC, sizeof(strC)); } lenA = strlen(strA); lenB = strlen(strB); if(lenA > lenB - 1){//这种情况就是A一定会比B要大，所以就可以用A去减掉B int start = 0; int ch = 0; int i, j; for(i = lenA - 1, j = lenB - 1; i >= 0 && j >= 1; i--, j--){ C[start] = strA[i] - strB[j] - ch; if(C[start] < 0) {C[start] = 10 + C[start]; ch = 1;} else ch = 0; start++; } while(i >= 0){ C[start++] = strA[i--] - '0' - ch; ch = 0; } for(int i = start-1; i >= 0; i--){ if(C[i]){ while(i>=0){ printf("%d", C[i]); i--; } printf("\n"); break; } } }//if else if(lenA == lenB - 1){ int ok; int start = 0; if(strncmp(strA, strB+1, lenA) == 0) {printf("0\n"); continue;} //之前把函数写错了，把strncmp写成了strncpy else if(strncmp(strA, strB+1, lenA) > 0) ok = 1; //strncpy是将后面字符串的前n个复制到前面 else ok = 0; if(ok){ //当A大于B时，就拿A去减掉B int start = 0; int ch = 0; int i, j; for(i = lenA - 1, j = lenB - 1; i >= 0 && j >= 1; i--, j--){ C[start] = strA[i] - strB[j]- ch; if(C[start] < 0) {C[start] = 10 + C[start]; ch = 1;} else ch = 0; start++; } for(int i = start-1; i >= 0; i--){ if(C[i]){ while(i>=0){ printf("%d", C[i]); i--; } printf("\n"); break; } } }//if else { int start = 0; int ch = 0; int i, j; for(i = lenA - 1, j = lenB - 1; i >= 0 && j >= 1; i--, j--){ C[start] = strB[j] - strA[i] - ch; if(C[start] < 0) {C[start] = 10 + C[start]; ch = 1;} else ch = 0; start++; } printf("-"); for(int i = start-1; i >= 0; i--){ if(C[i]){ while(i>=0){ printf("%d", C[i]); i--; }//while printf("\n"); break; }//if }//for }//else }//else if else {//这种情况就是指A的值的大小必定会小于B int start = 0; int ch = 0; int i, j; for(i = lenA - 1, j = lenB - 1; i >= 0 && j >= 1; i--, j--){ C[start] = strB[j] - strA[i] - ch; if(C[start] < 0) {C[start] = 10 + C[start]; ch = 1;} else ch = 0; start++; } while(j >= 1){ C[start++] = strB[j--] - '0' - ch; ch = 0; } printf("-"); for(int i = start-1; i >= 0; i--){ if(C[i]){ while(i>=0){ printf("%d", C[i]); i--; }//while printf("\n"); break; }//if }//for }//else }//else }//while（最外层的那个） return 0; }
Solution C++
#include <deque> #include <vector> #include <iostream> #include <string> #include <algorithm> #include <stdio.h> using namespace std; class DividedByZeroException {}; class BigInteger { private: vector<char> digits; bool sign; // true for positive, false for negitive void trim(); // remove zeros in tail, but if the value is 0, keep only one:) public: BigInteger(int); // construct with a int integer BigInteger(string&) ; BigInteger(); BigInteger (const BigInteger&); BigInteger operator=(const BigInteger& op2); BigInteger abs() const; BigInteger pow(int a); //binary operators friend BigInteger operator+=(BigInteger&,const BigInteger&); friend BigInteger operator-=(BigInteger&,const BigInteger&); friend BigInteger operator*=(BigInteger&,const BigInteger&); friend BigInteger operator/=(BigInteger&,const BigInteger&) throw(DividedByZeroException); friend BigInteger operator%=(BigInteger&,const BigInteger&) throw(DividedByZeroException); friend BigInteger operator+(const BigInteger&,const BigInteger&); friend BigInteger operator-(const BigInteger&,const BigInteger&); friend BigInteger operator*(const BigInteger&,const BigInteger&); friend BigInteger operator/(const BigInteger&,const BigInteger&) throw(DividedByZeroException); friend BigInteger operator%(const BigInteger&,const BigInteger&) throw(DividedByZeroException); //uniary operators friend BigInteger operator-(const BigInteger&); //negative friend BigInteger operator++(BigInteger&); //++v friend BigInteger operator++(BigInteger&,int); //v++ friend BigInteger operator--(BigInteger&); //--v friend BigInteger operator--(BigInteger&,int); //v-- friend bool operator>(const BigInteger&,const BigInteger&); friend bool operator<(const BigInteger&,const BigInteger&); friend bool operator==(const BigInteger&,const BigInteger&); friend bool operator!=(const BigInteger&,const BigInteger&); friend bool operator>=(const BigInteger&,const BigInteger&); friend bool operator<=(const BigInteger&,const BigInteger&); friend ostream& operator<<(ostream&,const BigInteger&); //print the BigInteger friend istream& operator>>(istream&, BigInteger&); // input the BigInteger public: static const BigInteger ZERO; static const BigInteger ONE; static const BigInteger TEN; }; // BigInteger.cpp const BigInteger BigInteger::ZERO=BigInteger(0); const BigInteger BigInteger::ONE =BigInteger(1); const BigInteger BigInteger::TEN =BigInteger(10); BigInteger::BigInteger() { sign=true; } BigInteger::BigInteger(int val){// construct with a int integer if (val >= 0) sign = true; else{ sign = false; val *= (-1); } do{ digits.push_back( (char)(val%10) ); val /= 10; } while ( val != 0 ); } BigInteger::BigInteger(string& def){ sign=true; for ( string::reverse_iterator iter = def.rbegin() ; iter < def.rend(); iter++){ char ch = (*iter); if (iter == def.rend()-1){ if ( ch == '+' ) break; if(ch == '-' ){ sign = false; break; } } digits.push_back( (char)((*iter) - '0' ) ); } trim(); } void BigInteger::trim(){ vector<char>::reverse_iterator iter = digits.rbegin(); while(!digits.empty() && (*iter) == 0){ digits.pop_back(); iter=digits.rbegin(); } if( digits.size()==0 ){ sign = true; digits.push_back(0); } } BigInteger::BigInteger(const BigInteger& op2){ sign = op2.sign; digits=op2.digits; } BigInteger BigInteger::operator=(const BigInteger& op2){ digits = op2.digits; sign = op2.sign; return (*this); } BigInteger BigInteger::abs() const { if(sign) return *this; else return -(*this); } BigInteger BigInteger::pow(int a) { BigInteger res(1); for(int i=0; i<a; i++) res*=(*this); return res; } //binary operators BigInteger operator+=(BigInteger& op1,const BigInteger& op2){ if( op1.sign == op2.sign ){ //只处理相同的符号的情况，异号的情况给-处理 vector<char>::iterator iter1; vector<char>::const_iterator iter2; iter1 = op1.digits.begin(); iter2 = op2.digits.begin(); char to_add = 0; //进位 while ( iter1 != op1.digits.end() && iter2 != op2.digits.end()){ (*iter1) = (*iter1) + (*iter2) + to_add; to_add = ((*iter1) > 9); // 大于9进一位 (*iter1) = (*iter1) % 10; iter1++; iter2++; } while ( iter1 != op1.digits.end() ){ // (*iter1) = (*iter1) + to_add; to_add = ( (*iter1) > 9 ); (*iter1) %= 10; iter1++; } while ( iter2 != op2.digits.end() ){ char val = (*iter2) + to_add; to_add = (val > 9) ; val %= 10; op1.digits.push_back(val); iter2++; } if( to_add != 0 ) op1.digits.push_back(to_add); return op1; } else{ if (op1.sign) return op1 -= (-op2); else return op1= op2 - (-op1); } } BigInteger operator-=(BigInteger& op1,const BigInteger& op2){ if( op1.sign == op2.sign ){ //只处理相同的符号的情况，异号的情况给+处理 if(op1.sign) { if(op1 < op2) // 2 - 3 return op1=-(op2 - op1); } else { if(-op1 > -op2) // (-3)-(-2) = -(3 - 2) return op1=-((-op1)-(-op2)); else // (-2)-(-3) = 3 - 2 return op1= (-op2) - (-op1); } vector<char>::iterator iter1; vector<char>::const_iterator iter2; iter1 = op1.digits.begin(); iter2 = op2.digits.begin(); char to_substract = 0; //借位 while ( iter1 != op1.digits.end() && iter2 != op2.digits.end()){ (*iter1) = (*iter1) - (*iter2) - to_substract; to_substract = 0; if( (*iter1) < 0 ){ to_substract=1; (*iter1) += 10; } iter1++; iter2++; } while ( iter1 != op1.digits.end() ){ (*iter1) = (*iter1) - to_substract; to_substract = 0; if( (*iter1) < 0 ){ to_substract=1; (*iter1) += 10; } else break; iter1++; } op1.trim(); return op1; } else{ if (op1 > BigInteger::ZERO) return op1 += (-op2); else return op1 = -(op2 + (-op1)); } } BigInteger operator*=(BigInteger& op1,const BigInteger& op2){ BigInteger result(0); if (op1 == BigInteger::ZERO || op2==BigInteger::ZERO) result = BigInteger::ZERO; else{ vector<char>::const_iterator iter2 = op2.digits.begin(); while( iter2 != op2.digits.end() ){ if(*iter2 != 0){ deque<char> temp(op1.digits.begin() , op1.digits.end()); char to_add = 0; deque<char>::iterator iter1 = temp.begin(); while( iter1 != temp.end() ){ (*iter1) *= (*iter2); (*iter1) += to_add; to_add = (*iter1) / 10; (*iter1) %= 10; iter1++; } if( to_add != 0) temp.push_back( to_add ); int num_of_zeros = iter2 - op2.digits.begin(); while( num_of_zeros--) temp.push_front(0); BigInteger temp2; temp2.digits.insert( temp2.digits.end() , temp.begin() , temp.end() ); temp2.trim(); result = result + temp2; } iter2++; } result.sign = ( (op1.sign && op2.sign) || (!op1.sign && !op2.sign) ); } op1 = result; return op1; } BigInteger operator/=(BigInteger& op1 , const BigInteger& op2 ) throw(DividedByZeroException) { if( op2 == BigInteger::ZERO ) throw DividedByZeroException(); BigInteger t1 = op1.abs(), t2 = op2.abs(); if ( t1 < t2 ){ op1 = BigInteger::ZERO; return op1; } //现在 t1 > t2 > 0 //只需将 t1/t2的结果交给result就可以了 deque<char> temp; vector<char>::reverse_iterator iter = t1.digits.rbegin(); BigInteger temp2(0); while( iter != t1.digits.rend() ){ temp2 = temp2 * BigInteger::TEN + BigInteger( (int)(*iter) ); char s = 0; while( temp2 >= t2 ){ temp2 = temp2 - t2; s = s + 1; } temp.push_front( s ); iter++; } op1.digits.clear(); op1.digits.insert( op1.digits.end() , temp.begin() , temp.end() ); op1.trim(); op1.sign = ( (op1.sign && op2.sign) || (!op1.sign && !op2.sign) ); return op1; } BigInteger operator%=(BigInteger& op1,const BigInteger& op2) throw(DividedByZeroException) { return op1 -= ((op1 / op2)*op2); } BigInteger operator+(const BigInteger& op1,const BigInteger& op2){ BigInteger temp(op1); temp += op2; return temp; } BigInteger operator-(const BigInteger& op1,const BigInteger& op2){ BigInteger temp(op1); temp -= op2; return temp; } BigInteger operator*(const BigInteger& op1,const BigInteger& op2){ BigInteger temp(op1); temp *= op2; return temp; } BigInteger operator/(const BigInteger& op1,const BigInteger& op2) throw(DividedByZeroException) { BigInteger temp(op1); temp /= op2; return temp; } BigInteger operator%(const BigInteger& op1,const BigInteger& op2) throw(DividedByZeroException) { BigInteger temp(op1); temp %= op2; return temp; } //uniary operators BigInteger operator-(const BigInteger& op){ //negative BigInteger temp = BigInteger(op); temp.sign = !temp.sign; return temp; } BigInteger operator++(BigInteger& op){ //++v op += BigInteger::ONE; return op; } BigInteger operator++(BigInteger& op,int x){ //v++ BigInteger temp(op); ++op; return temp; } BigInteger operator--(BigInteger& op){ //--v op -= BigInteger::ONE; return op; } BigInteger operator--(BigInteger& op,int x){ //v-- BigInteger temp(op); --op; return temp; } bool operator<(const BigInteger& op1,const BigInteger& op2){ if( op1.sign != op2.sign ) return !op1.sign; else{ if(op1.digits.size() != op2.digits.size()) return (op1.sign && op1.digits.size()<op2.digits.size()) || (!op1.sign && op1.digits.size()>op2.digits.size()); vector<char>::const_reverse_iterator iter1,iter2; iter1 = op1.digits.rbegin();iter2 = op2.digits.rbegin(); while( iter1 != op1.digits.rend() ){ if( op1.sign && *iter1 < *iter2 ) return true; if( op1.sign && *iter1 > *iter2 ) return false; if( !op1.sign && *iter1 > *iter2 ) return true; if( !op1.sign && *iter1 < *iter2 ) return false; iter1++; iter2++; } return false; } } bool operator==(const BigInteger& op1,const BigInteger& op2){ if( op1.sign != op2.sign || op1.digits.size() != op2.digits.size() ) return false; vector<char>::const_iterator iter1,iter2; iter1 = op1.digits.begin(); iter2 = op2.digits.begin(); while( iter1!= op1.digits.end() ){ if( *iter1 != *iter2 ) return false; iter1++; iter2++; } return true; } bool operator!=(const BigInteger& op1,const BigInteger& op2){ return !(op1==op2); } bool operator>=(const BigInteger& op1,const BigInteger& op2){ return (op1>op2) || (op1==op2); } bool operator<=(const BigInteger& op1,const BigInteger& op2){ return (op1<op2) || (op1==op2); } bool operator>(const BigInteger& op1,const BigInteger& op2){ return !(op1<=op2); } ostream& operator<<(ostream& stream,const BigInteger& val){ //print the BigInteger if (!val.sign) stream << "-"; for ( vector<char>::const_reverse_iterator iter = val.digits.rbegin(); iter != val.digits.rend() ; iter++) stream << (char)((*iter) + '0'); return stream; } istream& operator>>(istream& stream, BigInteger& val){ //Input the BigInteger string str; stream >> str; val=BigInteger(str); return stream; } int main() { // freopen("1.txt","r",stdin); // freopen("2.txt","w",stdout); string a,b; while(cin>>a>>b) { BigInteger c = BigInteger(a); BigInteger d = BigInteger(b); cout<< c+d <<endl; } }
Hint
Bazinga !!!