hdu 1250 Hat's Fibonacci
来源:互联网 发布:美国的种族歧视 知乎 编辑:程序博客网 时间:2024/06/10 22:23
原题链接:http://acm.hdu.edu.cn/showproblem.php?pid=1250
大数加法模板题。。
#include<algorithm>#include<iostream>#include<cstdlib>#include<cstring>#include<cassert>#include<cstdio>#include<vector>#include<string>#include<map>#include<set>using std::cin;using std::max;using std::cout;using std::endl;using std::string;using std::istream;using std::ostream;#define sz(c) (int)(c).size()#define all(c) (c).begin(), (c).end()#define iter(c) decltype((c).begin())#define cls(arr,val) memset(arr,val,sizeof(arr))#define cpresent(c, e) (find(all(c), (e)) != (c).end())#define rep(i, n) for (int i = 0; i < (int)(n); i++)#define fork(i, k, n) for (int i = (int)k; i <= (int)n; i++)#define tr(c, i) for (iter(c) i = (c).begin(); i != (c).end(); ++i)#define pb(e) push_back(e)#define mp(a, b) make_pair(a, b)struct BigN { typedef unsigned long long ull; static const int Max_N = 2010; int len, data[Max_N]; BigN() { memset(data, 0, sizeof(data)), len = 0; } BigN(const int num) { memset(data, 0, sizeof(data)); *this = num; } BigN(const char *num) { memset(data, 0, sizeof(data)); *this = num; } void clear() { len = 0, memset(data, 0, sizeof(data)); } BigN& clean(){ while (len > 1 && !data[len - 1]) len--; return *this; } string str() const { string res = ""; for (int i = len - 1; ~i; i--) res += (char)(data[i] + '0'); if (res == "") res = "0"; res.reserve(); return res; } BigN operator = (const int num) { int j = 0, i = num; do data[j++] = i % 10; while (i /= 10); len = j; return *this; } BigN operator = (const char *num) { len = strlen(num); for (int i = 0; i < len; i++) data[i] = num[len - i - 1] - '0'; return *this; } BigN operator + (const BigN &x) const { BigN res; int n = max(len, x.len) + 1; for (int i = 0, g = 0; i < n; i++) { int c = data[i] + x.data[i] + g; res.data[res.len++] = c % 10; g = c / 10; } return res.clean(); } BigN operator * (const BigN &x) const { BigN res; int n = x.len; res.len = n + len; for (int i = 0; i < len; i++) { for (int j = 0, g = 0; j < n; j++) { res.data[i + j] += data[i] * x.data[j]; } } for (int i = 0; i < res.len - 1; i++) { res.data[i + 1] += res.data[i] / 10; res.data[i] %= 10; } return res.clean(); } BigN operator * (const int num) const { BigN res; res.len = len + 1; for (int i = 0, g = 0; i < len; i++) res.data[i] *= num; for (int i = 0; i < res.len - 1; i++) { res.data[i + 1] += res.data[i] / 10; res.data[i] %= 10; } return res.clean(); } BigN operator - (const BigN &x) const { assert(x <= *this); BigN res; for (int i = 0, g = 0; i < len; i++) { int c = data[i] - g; if (i < x.len) c -= x.data[i]; if (c >= 0) g = 0; else g = 1, c += 10; res.data[res.len++] = c; } return res.clean(); } BigN operator / (const BigN &x) const { BigN res, f = 0; for (int i = len - 1; ~i; i--) { f *= 10; f.data[0] = data[i]; while (f >= x) { f -= x; res.data[i]++; } } res.len = len; return res.clean(); } BigN operator % (const BigN &x) { BigN res = *this / x; res = *this - res * x; return res; } BigN operator += (const BigN &x) { return *this = *this + x; } BigN operator *= (const BigN &x) { return *this = *this * x; } BigN operator -= (const BigN &x) { return *this = *this - x; } BigN operator /= (const BigN &x) { return *this = *this / x; } BigN operator %= (const BigN &x) { return *this = *this % x; } bool operator < (const BigN &x) const { if (len != x.len) return len < x.len; for (int i = len - 1; ~i; i--) { if (data[i] != x.data[i]) return data[i] < x.data[i]; } return false; } bool operator >(const BigN &x) const { return x < *this; } bool operator<=(const BigN &x) const { return !(x < *this); } bool operator>=(const BigN &x) const { return !(*this < x); } bool operator!=(const BigN &x) const { return x < *this || *this < x; } bool operator==(const BigN &x) const { return !(x < *this) && !(x > *this); } friend istream& operator >> (istream &in, BigN &x) { string src; in >> src; x = src.c_str(); return in; } friend ostream& operator << (ostream &out, const BigN &x) { out << x.str(); return out; }}A[5];void solve(int n) { fork(i, 1, 4) A[i] = 1; if (n < 5) cout << A[n] << endl; else { int x = 1; fork(i, 5, n) { A[0] = A[1] + A[2] + A[3] + A[4]; A[x] = A[0]; if (++x == 5) x = 1; } cout << A[0] << endl; } rep(i, 5) A[i].clear();}int main() {#ifdef LOCAL freopen("in.txt", "r", stdin); freopen("out.txt", "w+", stdout);#endif std::ios::sync_with_stdio(false); int n; while (cin >> n) { solve(n); } return 0;}
0 0
- hdu 1250 Hat's Fibonacci
- HDU 1250 Hat's Fibonacci
- Hdu 1250 Hat's Fibonacci
- hdu 1250 Hat's Fibonacci
- HDU 1250-Hat's Fibonacci
- hdu 1250 Hat's Fibonacci
- hdu-1250-Hat's Fibonacci
- Hdu---Hat's Fibonacci---1250
- hdu 1250 Hat's Fibonacci
- hdu 1250 Hat's Fibonacci
- hdu 1250 Hat's Fibonacci
- hdu 1250 Hat's Fibonacci
- HDU 1250: Hat's Fibonacci
- hdu 1250 Hat's Fibonacci
- hdu 1250 Hat's Fibonacci
- hdu-1250-Hat's Fibonacci
- HDU 1250 Hat's Fibonacci
- HDU 1250 Hat's Fibonacci
- 常用开发语言小技巧(一)
- HTML5摇一摇
- pandas学习(三)
- 简单尝试 Cocos2d-x 中 Lambda 表达式
- IOS开发入门书籍
- hdu 1250 Hat's Fibonacci
- 10110 - Light, more light
- C++ 虚函数 剖析
- 【leetcode】Gas Station
- java并发编程第四章 线程执行器(1)
- 有深度,面试有用的题
- java基础 第5章 隐藏实施过程
- C++:模板实参推断及引用折叠
- boost内存管理机制