UVA 763 - Fibinary Numbers(高精度斐波那契 + 高精度模板)
2015-03-03 23:29
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题意:
给2个斐波那契进制数。求他们的和,输出也要是斐波那契进制数,并且是不能有相邻的1的表示方法思路:
高精度,先把两个二进制数字转化为10进制数字相加,每次找到第一个大于sum的f[i]数字,将sum -= f[i],同时将ans的当前位变为1AC代码
[code]#include <cstdio> #include <cstring> #include <cstdlib> #include <iostream> #include <algorithm> typedef long long ll; using namespace std; #define MAXN 9999 #define MAXSIZE 10 #define DLEN 4 class BigNum { private: int a[500]; int len; public: BigNum() { len = 1; memset(a,0,sizeof(a)); } BigNum(const ll); BigNum(const int); BigNum(const char*); BigNum(const BigNum &); 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 BigNum & T) const; bool operator==(const BigNum & T) const; bool operator>(const int & t)const; //大数和一个int类型的变量的大小比较 bool operator<(const int & t) const; bool operator==(const int & t) const; void print(); //输出大数 }; bool BigNum::operator==(const BigNum & T) const { return !(*this > T) && !(T > *this); } bool BigNum::operator==(const int &t) const { BigNum T = BigNum(t); return *this == T; } bool BigNum::operator<(const BigNum & T) const { return T > *this; } bool BigNum::operator<(const int &t) const { return BigNum(t) > *this; } BigNum::BigNum(const ll b) { ll 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 int b) { 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) { int t,k,index,l; memset(a,0,sizeof(a)); l = strlen(s); len=l / DLEN; if(l % DLEN) len++; index=0; for(int 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) { memset(a,0,sizeof(a)); for(int i = 0 ; i < len; i++) a[i] = T.a[i]; } BigNum & BigNum::operator=(const BigNum & n) { len = n.len; memset(a,0,sizeof(a)); for(int i = 0; i < len; i++) a[i] = n.a[i]; return *this; } istream& operator>>(istream & in, BigNum & b) { 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) { 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 d=0; for (int 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() { printf("%d", a[len-1]); for (int i = len-2; i >= 0; --i) { printf("%04d", a[i]); } } const int N = 105; char fstr , sstr ; BigNum f ; void init() { f[0] = BigNum(0); f[1] = BigNum(1); f[2] = BigNum(2); for(int i = 3; i < N; i++) { f[i] = f[i-1] + f[i-2]; } } int findStar(BigNum num) { for(int i = 1; i < 105; i++) { if(f[i] > num) return i - 1; } return 0; } int main() { int cas = 0; init(); while(scanf("%s%s", fstr, sstr) != EOF) { if(cas++) puts(""); BigNum sum = BigNum(0); int len1 = strlen(fstr) , len2 = strlen(sstr); for(int i = 0; i < len1; i++) { if(fstr[i] == '1') { sum = sum + f[len1 - i]; } } for(int i = 0; i < len2; i++) { if(sstr[i] == '1') { sum = sum + f[len2 - i]; } } int idx = findStar(sum); string ans = ""; for(int i = idx; i >= 1; i--) { if(f[i] < sum || f[i] == sum) { sum = sum - f[i]; ans += "1"; }else { ans += "0"; } } if(idx == 0) ans += "0"; printf("%s\n", ans.c_str()); } return 0; }
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