• UVA 10334 Ray Through Glasses


    自己手动画了第三项发现f[3]=5;就猜斐波那契了。实际上光线分为两种距离外界有2面玻璃,1面玻璃 其分别时n-1次反射,n-2次反射形成的

    故推出斐波那契。 手动一些f1,f2,f3就OK

    #include <map>
    #include <set>
    #include <list>
    #include <cmath>
    #include <ctime>
    #include <deque>
    #include <stack>
    #include <queue>
    #include <cctype>
    #include <cstdio>
    #include <string>
    #include <vector>
    #include <climits>
    #include <cstdlib>
    #include <cstring>
    #include <iostream>
    #include <algorithm>
    #define LL long long
    #define PI 3.1415926535897932626
    using namespace std;
    int gcd(int a, int b) {return a % b == 0 ? b : gcd(b, a % b);}
    const int numlen=1010;
    struct bign {
        int len, s[numlen];
        bign() {
            memset(s, 0, sizeof(s));
            len = 1;
        }
        bign(int num) { *this = num; }
        bign(const char *num) { *this = num; }
        bign operator = (const int num) {
            char s[numlen];
            sprintf(s, "%d", num);
            *this = s;
            return *this;
        }
        bign operator = (const char *num) {
            len = strlen(num);
            while(len > 1 && num[0] == '0') num++, len--;
            for(int i = 0;i < len; i++) s[i] = num[len-i-1] - '0';
            return *this;
        }
    
        void deal() {
            while(len > 1 && !s[len-1]) len--;
        }
    
        bign operator + (const bign &a) const {
            bign ret;
            ret.len = 0;
            int top = max(len, a.len) , add = 0;
            for(int i = 0;add || i < top; i++) {
                int now = add;
                if(i < len) now += s[i];
                if(i < a.len)   now += a.s[i];
                ret.s[ret.len++] = now%10;
                add = now/10;
            }
            return ret;
        }
        bign operator - (const bign &a) const {
            bign ret;
            ret.len = 0;
            int cal = 0;
            for(int i = 0;i < len; i++) {
                int now = s[i] - cal;
                if(i < a.len)   now -= a.s[i];
                if(now >= 0)    cal = 0;
                else {
                    cal = 1; now += 10;
                }
                ret.s[ret.len++] = now;
            }
            ret.deal();
            return ret;
        }
        bign operator * (const bign &a) const {
            bign ret;
            ret.len = len + a.len;
            for(int i = 0;i < len; i++) {
                for(int j = 0;j < a.len; j++)
                    ret.s[i+j] += s[i]*a.s[j];
            }
            for(int i = 0;i < ret.len; i++) {
                ret.s[i+1] += ret.s[i]/10;
                ret.s[i] %= 10;
            }
            ret.deal();
            return ret;
        }
    
        bign operator * (const int num) {
    //        printf("num = %d
    ", num);
            bign ret;
            ret.len = 0;
            int bb = 0;
            for(int i = 0;i < len; i++) {
                int now = bb + s[i]*num;
                ret.s[ret.len++] = now%10;
                bb = now/10;
            }
            while(bb) {
                ret.s[ret.len++] = bb % 10;
                bb /= 10;
            }
            ret.deal();
            return ret;
        }
    
        bign operator / (const bign &a) const {
            bign ret, cur = 0;
            ret.len = len;
            for(int i = len-1;i >= 0; i--) {
                cur = cur*10;
                cur.s[0] = s[i];
                while(cur >= a) {
                    cur -= a;
                    ret.s[i]++;
                }
            }
            ret.deal();
            return ret;
        }
    
        bign operator % (const bign &a) const {
            bign b = *this / a;
            return *this - b*a;
        }
    
        bign operator += (const bign &a) { *this = *this + a; return *this; }
        bign operator -= (const bign &a) { *this = *this - a; return *this; }
        bign operator *= (const bign &a) { *this = *this * a; return *this; }
        bign operator /= (const bign &a) { *this = *this / a; return *this; }
        bign operator %= (const bign &a) { *this = *this % a; return *this; }
    
        bool operator < (const bign &a) const {
            if(len != a.len)    return len < a.len;
            for(int i = len-1;i >= 0; i--) if(s[i] != a.s[i])
                return s[i] < a.s[i];
            return false;
        }
        bool operator > (const bign &a) const  { return a < *this; }
        bool operator <= (const bign &a) const { return !(*this > a); }
        bool operator >= (const bign &a) const { return !(*this < a); }
        bool operator == (const bign &a) const { return !(*this > a || *this < a); }
        bool operator != (const bign &a) const { return *this > a || *this < a; }
    
        string str() const {
            string ret = "";
            for(int i = 0;i < len; i++) ret = char(s[i] + '0') + ret;
            return ret;
        }
    };
    istream& operator >> (istream &in, bign &x) {
        string s;
        in >> s;
        x = s.c_str();
        return in;
    }
    ostream& operator << (ostream &out, const bign &x) {
        out << x.str();
        return out;
    }
    bign res[1005];
    void init()
    {
        res[0]=1;res[1]=2;
        for (int i=2;i<1005;i++)
            res[i]=res[i-1]+res[i-2];
    }
    int main()
    {
        init();
        int N;
        while (scanf("%d",&N)!=EOF)
        cout<<res[N]<<endl;
        return 0;
    }
  • 相关阅读:
    部分模块
    正则表达式和re模块
    软件开发的目录规范
    模块以及模块的导入
    迭代器、可迭代对象、迭代器对象、生成器、生成器表达式和相关的面试题
    内置方法补充
    递归、二分法、匿名函数和部分内置方法
    python的闭包和装饰器
    函数名本质、函数命名空间和作用域链
    函数简介和函数的结构分析
  • 原文地址:https://www.cnblogs.com/Commence/p/3977788.html
Copyright © 2020-2023  润新知