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POJ 2084
第一题组合数学题。可以使用递推,设1与其他各数分别连边,假设N=3;若1-4,则圆分成两部分计数,此时可以利用乘法原理。(高精度)
#include <cstdio>#include <cstring>#include <iostream>#include <string>using namespace std;const int maxn = 200;struct bign { int len, s[maxn]; bign() { memset(s, 0, sizeof(s)); len = 1; } bign(int num) { *this = num; } bign(const char* num) { *this = num; } bign operator =(int num) { //直接以整数赋值 char s[maxn]; sprintf(s, "%d", num); *this = s; return *this; } bign operator =(const char* num) { //以字符串赋值 len = strlen(num); for(int i = 0; i < len; i++) s[i] = num[len - i - 1] - ‘0‘; return *this; } string str() const { //将bign转化成字符串 string res = ""; for(int i = 0; i < len; i++) res = (char) (s[i] + ‘0‘) + res; if(res == "") res = "0"; return res; } bign operator +(const bign& b) const { //重载+号运算 bign c; c.len = 0; for(int i = 0, g = 0; g || i < max(len, b.len); i++) { int x = g; if(i < len) x += s[i]; if(i < b.len) x += b.s[i]; c.s[c.len++] = x % 10; g = x / 10; } return c; } void clean() { //去掉前到0 while(len > 1 && !s[len - 1]) len--; } bign operator *(const bign& b) { //重载*号运算 bign c; c.len = len + b.len; for(int i = 0; i < len; i++) for(int j = 0; j < b.len; j++) c.s[i + j] += s[i] * b.s[j]; for(int i = 0; i < c.len - 1; i++) { c.s[i + 1] += c.s[i] / 10; c.s[i] %= 10; } c.clean(); return c; } bign operator -(const bign& b) { //重载-号运算 bign c; c.len = 0; for(int i = 0, g = 0; i < len; i++) { int x = s[i] - g; if(i < b.len) x -= b.s[i]; if(x >= 0) g = 0; else { g = 1; x += 10; } c.s[c.len++] = x; } c.clean(); return c; } bool operator <(const bign& b) const { //重载<号运算 if(len != b.len) return len < b.len; for(int i = len - 1; i >= 0; i--) if(s[i] != b.s[i]) return s[i] < b.s[i]; return false; } bool operator >(const bign& b) const { //重载>号运算 return b < *this; } bool operator <=(const bign& b) { //重载<=号运算 return !(b > *this); } bool operator ==(const bign& b) { //重载>=号运算 return !(b < *this) && !(*this < b); } bign operator +=(const bign& b) { //重载+=号运算 *this = *this + b; return *this; }};istream& operator >>(istream &in, bign& x) { //重载输入运算符 string s; in >> s; x = s.c_str(); return in;}ostream& operator <<(ostream &out, const bign& x) { //重载输出运算符 out << x.str(); return out;}bign ans[210];void initial(){ ans[2]=1; ans[4]=2; for(int i=5;i<=200;i++){ ans[i]=ans[i-2]+ans[i-2]; for(int j=3;j<i;j++){ ans[i]=ans[i]+ans[j-2]*ans[i-j]; } }}int main(){ initial(); int n; while(scanf("%d",&n),n!=-1){ cout<<ans[2*n]<<endl; } return 0;}
POJ 2084
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