UVA 10814 - Simplifying Fractions UVA • Sep 1, 2020 Problem PDFSolution:#include #include #include #include #include #include using namespace std; const int MAXD = 10000, DIG = 9, BASE = 1000000000; const unsigned long long BOUND = numeric_limits :: max () - (unsigned long long) BASE * BASE; class BigInteger{ private: int digits[MAXD]; int D; public: friend ostream &operator<<(ostream &out,BigInteger &c); inline void trim() { while(D > 1 && digits[D-1] == 0 ) D--; } inline void dealint(long long x) { memset(digits,0,sizeof(digits)); D = 0; do{ digits[D++] = x % BASE; x /= BASE; }while(x > 0); } inline void dealstr(char *s) { memset(digits,0,sizeof(digits)); int len = strlen(s),first = (len + DIG -1)%DIG + 1; D = (len+DIG-1)/DIG; for(int i = 0;i < first;i++) digits[D-1] = digits[D-1]*10 + s[i] - '0'; for(int i = first, d = D-2; i < len;i+=DIG,d--) for(int j = i;j < i+DIG;j++) digits[d] = digits[d]*10 + s[j]-'0'; trim(); } inline char *print() { trim(); char *cdigits = new char[DIG * D + 1]; int pos = 0,d = digits[D-1]; do{ cdigits[pos++] = d % 10 + '0'; d/=10; }while(d > 0); reverse(cdigits,cdigits+pos); for(int i = D - 2;i >= 0;i--,pos += DIG) for(int j = DIG-1,t = digits[i];j >= 0;j--) { cdigits[pos+j] = t%10 + '0'; t /= 10; } cdigits[pos] = '\0'; return cdigits; } BigInteger(){dealint(0);} BigInteger(long long x){ dealint(x); } BigInteger(int x){ dealint(x); } BigInteger(char *s){ dealstr(s); } inline bool operator < (const BigInteger &o) const { if(D != o.D) return D < o.D; for(int i = D-1;i>=0;i--) if(digits[i] != o.digits[i]) return digits[i] < o.digits[i]; return false; //equal } bool operator > (const BigInteger & o)const {return o < *this;} bool operator <= (const BigInteger & o)const {return !(o < *this);} bool operator >= (const BigInteger & o)const {return !(*this < o);} bool operator != (const BigInteger & o)const {return o < *this || *this < o;} bool operator == (const BigInteger & o)const {return !(o < *this) && !(*this < o);} BigInteger &operator++() { *this = *this + 1; return *this; } BigInteger operator ++(int) { BigInteger old = *this; ++(*this); return old; } inline BigInteger operator << (int p) const { BigInteger temp; temp.D = D + p; for (int i = 0; i < D; i++) temp.digits [i + p] = digits [i]; for (int i = 0; i < p; i++) temp.digits [i] = 0; return temp; } inline BigInteger operator >> (int p) const { BigInteger temp; temp.D = D - p; for (int i = 0; i < D - p; i++) temp.digits [i] = digits [i + p]; for (int i = D - p; i < D; i++) temp.digits [i] = 0; return temp; } BigInteger &operator += (const BigInteger &b) { *this = *this + b; return *this; } BigInteger &operator -= (const BigInteger &b) { *this = *this - b; return *this; } BigInteger &operator *= (const BigInteger &b) { *this = *this * b; return *this; } BigInteger &operator /= (const BigInteger &b) { *this = *this / b; return *this; } BigInteger &operator %= (const BigInteger &b) { *this = *this % b; return *this; } inline BigInteger operator + (const BigInteger &o) const { BigInteger sum = o; int carry = 0; for (sum.D = 0; sum.D < D || carry > 0; sum.D++) { sum.digits [sum.D] += (sum.D < D ? digits [sum.D] : 0) + carry; if (sum.digits [sum.D] >= BASE) { sum.digits [sum.D] -= BASE; carry = 1; } else carry = 0; } sum.D = max (sum.D, o.D); sum.trim (); return sum; } inline BigInteger operator - (const BigInteger &o) const { BigInteger diff = *this; for (int i = 0, carry = 0; i < o.D || carry > 0; i++) { diff.digits [i] -= (i < o.D ? o.digits [i] : 0) + carry; if (diff.digits [i] < 0) { diff.digits [i] += BASE; carry = 1; } else carry = 0; } diff.trim (); return diff; } inline BigInteger operator * (const BigInteger &o) const { BigInteger prod = 0; unsigned long long sum = 0, carry = 0; for (prod.D = 0; prod.D < D + o.D - 1 || carry > 0; prod.D++) { sum = carry % BASE; carry /= BASE; for (int j = max (prod.D - o.D + 1, 0); j <= min (D - 1, prod.D); j++) { sum += (unsigned long long) digits [j] * o.digits [prod.D - j]; if (sum >= BOUND) { carry += sum / BASE; sum %= BASE; } } carry += sum / BASE; prod.digits [prod.D] = sum % BASE; } prod.trim (); return prod; } inline BigInteger range (int a, int b) const { BigInteger temp = 0; temp.D = b - a; for (int i = 0; i < temp.D; i++) temp.digits [i] = digits [i + a]; return temp; } inline double double_div (const BigInteger &o) const { double val = 0, oval = 0; int num = 0, onum = 0; for (int i = D - 1; i >= max (D - 3, 0); i--, num++) val = val * BASE + digits [i]; for (int i = o.D - 1; i >= max (o.D - 3, 0); i--, onum++) oval = oval * BASE + o.digits [i]; return val / oval * (D - num > o.D - onum ? BASE : 1); } inline pair divmod (const BigInteger &o) const { BigInteger quot = 0, rem = *this, temp; for (int i = D - o.D; i >= 0; i--) { temp = rem.range (i, rem.D); int div = (int) temp.double_div (o); BigInteger mult = o * div; while (div > 0 && temp < mult) { mult = mult - o; div--; } while (div + 1 < BASE && !(temp < mult + o)) { mult = mult + o; div++; } rem = rem - (o * div << i); if (div > 0) { quot.digits [i] = div; quot.D = max (quot.D, i + 1); } } quot.trim (); rem.trim (); return make_pair (quot, rem); } inline BigInteger operator / (const BigInteger &o) const { return divmod (o).first; } inline BigInteger operator % (const BigInteger &o) const { return divmod (o).second; } inline BigInteger power (int exp) const { BigInteger p = 1, temp = *this; while (exp > 0) { if (exp & 1) p = p * temp; if (exp > 1) temp = temp * temp; exp >>= 1; } return p; } inline BigInteger factorial() const { BigInteger ans = 1, num = *this; if (num == 0 || num == 1) return ans; while (!(num < 0 || num == 0)) { ans = ans * num; num = num - 1; } return ans; } }; ostream &operator<<(ostream &out, BigInteger &c){ out<> (istream &in,BigInteger &c) { char s[10000]; in>>s; c = s; return in; } BigInteger gcd(BigInteger a,BigInteger b) { if (b==0) return a; return gcd(b, a%b); } BigInteger a,b,c; int main() { char k; int T; scanf("%d",&T); while(T--) { cin>>a>>k>>b; BigInteger p=gcd(a,b); a/=p,b/=p; cout<https://github.com/Shipu/OnlineJudgeProblemSolutionWithCPlusPlus/tree/master/uva/10814/10814.cpp Tags UVA Competitive Programming Shipu Ahamed Recommended for you UVA UVA 913 - Joana and the Odd Numbers 4 years ago • 1 min read UVA UVA 871 - Counting Cells in a Blob 4 years ago • 3 min read UVA UVA 866 - Intersecting Line Segments 4 years ago • 3 min read