斐波拉契高精度(洛谷1255)

分析：第n次的台阶数为dp[n]，则dp[n]=dp[n-1]+dp[n-2];

  1 //
  2 //  main.cpp
  3 //  1601
  4 //
  5 //  Created by wanghan on 16/10/12.
  6 //  Copyright © 2016年 wanghan. All rights reserved.
  7 //
  8
  9 #include<cstdio>
 10 #include<cstring>
 11 #include<vector>
 12 #include<iostream>
 13 #include<set>
 14 #include<map>
 15 #include<cassert>
 16 using namespace std;
 17
 18 using namespace std;
 19
 20 struct BigInteger {
 21     typedef unsigned long long LL;
 22
 23     static const int BASE = 100000000;
 24     static const int WIDTH = 8;
 25     vector<int> s;
 26
 27     BigInteger& clean(){while(!s.back()&&s.size()>1)s.pop_back(); return *this;}
 28     BigInteger(LL num = 0) {*this = num;}
 29     BigInteger(string s) {*this = s;}
 30     BigInteger& operator = (long long num) {
 31         s.clear();
 32         do {
 33             s.push_back(num % BASE);
 34             num /= BASE;
 35         } while (num > 0);
 36         return *this;
 37     }
 38     BigInteger& operator = (const string& str) {
 39         s.clear();
 40         int x, len = (str.length() - 1) / WIDTH + 1;
 41         for (int i = 0; i < len; i++) {
 42             int end = str.length() - i*WIDTH;
 43             int start = max(0, end - WIDTH);
 44             sscanf(str.substr(start,end-start).c_str(), "%d", &x);
 45             s.push_back(x);
 46         }
 47         return (*this).clean();
 48     }
 49
 50     BigInteger operator + (const BigInteger& b) const {
 51         BigInteger c; c.s.clear();
 52         for (int i = 0, g = 0; ; i++) {
 53             if (g == 0 && i >= s.size() && i >= b.s.size()) break;
 54             int x = g;
 55             if (i < s.size()) x += s[i];
 56             if (i < b.s.size()) x += b.s[i];
 57             c.s.push_back(x % BASE);
 58             g = x / BASE;
 59         }
 60         return c;
 61     }
 62     BigInteger operator - (const BigInteger& b) const {
 63         assert(b <= *this); // 减数不能大于被减数
 64         BigInteger c; c.s.clear();
 65         for (int i = 0, g = 0; ; i++) {
 66             if (g == 0 && i >= s.size() && i >= b.s.size()) break;
 67             int x = s[i] + g;
 68             if (i < b.s.size()) x -= b.s[i];
 69             if (x < 0) {g = -1; x += BASE;} else g = 0;
 70             c.s.push_back(x);
 71         }
 72         return c.clean();
 73     }
 74     BigInteger operator * (const BigInteger& b) const {
 75         int i, j; LL g;
 76         vector<LL> v(s.size()+b.s.size(), 0);
 77         BigInteger c; c.s.clear();
 78         for(i=0;i<s.size();i++) for(j=0;j<b.s.size();j++) v[i+j]+=LL(s[i])*b.s[j];
 79         for (i = 0, g = 0; ; i++) {
 80             if (g ==0 && i >= v.size()) break;
 81             LL x = v[i] + g;
 82             c.s.push_back(x % BASE);
 83             g = x / BASE;
 84         }
 85         return c.clean();
 86     }
 87     BigInteger operator / (const BigInteger& b) const {
 88         assert(b > 0);  // 除数必须大于0
 89         BigInteger c = *this;       // 商:主要是让c.s和(*this).s的vector一样大
 90         BigInteger m;               // 余数:初始化为0
 91         for (int i = s.size()-1; i >= 0; i--) {
 92             m = m*BASE + s[i];
 93             c.s[i] = bsearch(b, m);
 94             m -= b*c.s[i];
 95         }
 96         return c.clean();
 97     }
 98     BigInteger operator % (const BigInteger& b) const { //方法与除法相同
 99         BigInteger c = *this;
100         BigInteger m;
101         for (int i = s.size()-1; i >= 0; i--) {
102             m = m*BASE + s[i];
103             c.s[i] = bsearch(b, m);
104             m -= b*c.s[i];
105         }
106         return m;
107     }
108     // 二分法找出满足bx<=m的最大的x
109     int bsearch(const BigInteger& b, const BigInteger& m) const{
110         int L = 0, R = BASE-1, x;
111         while (1) {
112             x = (L+R)>>1;
113             if (b*x<=m) {if (b*(x+1)>m) return x; else L = x;}
114             else R = x;
115         }
116     }
117     BigInteger& operator += (const BigInteger& b) {*this = *this + b; return *this;}
118     BigInteger& operator -= (const BigInteger& b) {*this = *this - b; return *this;}
119     BigInteger& operator *= (const BigInteger& b) {*this = *this * b; return *this;}
120     BigInteger& operator /= (const BigInteger& b) {*this = *this / b; return *this;}
121     BigInteger& operator %= (const BigInteger& b) {*this = *this % b; return *this;}
122
123     bool operator < (const BigInteger& b) const {
124         if (s.size() != b.s.size()) return s.size() < b.s.size();
125         for (int i = s.size()-1; i >= 0; i--)
126             if (s[i] != b.s[i]) return s[i] < b.s[i];
127         return false;
128     }
129     bool operator >(const BigInteger& b) const{return b < *this;}
130     bool operator<=(const BigInteger& b) const{return !(b < *this);}
131     bool operator>=(const BigInteger& b) const{return !(*this < b);}
132     bool operator!=(const BigInteger& b) const{return b < *this || *this < b;}
133     bool operator==(const BigInteger& b) const{return !(b < *this) && !(b > *this);}
134 };
135
136 ostream& operator << (ostream& out, const BigInteger& x) {
137     out << x.s.back();
138     for (int i = x.s.size()-2; i >= 0; i--) {
139         char buf[20];
140         sprintf(buf, "%08d", x.s[i]);
141         for (int j = 0; j < strlen(buf); j++) out << buf[j];
142     }
143     return out;
144 }
145
146 istream& operator >> (istream& in, BigInteger& x) {
147     string s;
148     if (!(in >> s)) return in;
149     x = s;
150     return in;
151 }
152 set<BigInteger> s;
153 map<BigInteger, int> m;
154
155 int main() {
156     BigInteger dp[5050];
157     int n;
158     while(cin>>n)
159     {
160         for(int i=0;i<5050;++i)
161             dp[i]=BigInteger(0);
162         dp[1]=BigInteger(1);
163         dp[2]=BigInteger(2);
164         if(n<=2){
165             cout<<dp[n]<<endl;
166             continue;
167         }
168         for(int i=3;i<=n;i++)
169             dp[i]=dp[i-1]+dp[i-2];
170         cout<<dp[n]<<endl;
171     }
172     return 0;
173 }