https://www.luogu.org/problem/show?pid=T2379
题目背景
宁静祥和的原始森林,空气清新,万物复苏。小Z和小V在林间追跑,快乐非常。
题目描述
小Z好吃懒做, 虽然很胖可是却没有小V健壮, 很是不服气。就提出要求和小V比体重,谁重谁当“头头” 。虽然知道小V的体重是 a、小Z的体重是 b,如果a 小于 b,小Z做头头,否则,小V做头头。但小V还是答应了这个要求。聪明的小V称体重前偷偷地喝了 c 重量的水。请你判断比体重后谁是头头?
如果小V是头头输出小V及他的重量,如果小Z是头头输出小Z以及他的重量。
输入输出格式
输入格式:
一行,三个数,空格分开,表示 a、b 和 c。
输出格式:
两行,分别是一个字串以及一个数,
第一行:头头是小V还是小Z。
第二行:小V的重量或小Z的重量。
输入输出样例
输入样例#1:
5 2 2
输出样例#1:
little v height:7
输入样例#2:
52.2 2 0
输出样例#2:
little v height:52.2
说明
Hint#1:a,b,c范围在200位以内
Hint#2:重量相同时,小V仍然是头头
Hint#3:若不是纯数字,输出:wrong
Hint#4:a,b,c可能是小数
Hint#5:a,b,c都是高精度(别傻傻的用高精+单精)
#include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #include <string> #include <algorithm> using namespace std; const int MAXN = 410; 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 = (const int num) { char s[MAXN]; sprintf(s, "%d", num); *this = s; return *this; } bign operator = (const char *num) { for(int i = 0; num[i] == '0'; num++) ; //去前导0 len = strlen(num); for(int i = 0; i < len; i++) s[i] = num[len-i-1] - '0'; return *this; } bign operator + (const bign &b) const //+ { 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; } bign operator += (const bign &b) { *this = *this + b; return *this; } void clean() { while(len > 1 && !s[len-1]) len--; } bign operator * (const bign &b) //* { 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; i++) { c.s[i+1] += c.s[i]/10; c.s[i] %= 10; } c.clean(); return c; } bign operator *= (const bign &b) { *this = *this * b; return *this; } bign operator - (const bign &b) { 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; } bign operator -= (const bign &b) { *this = *this - b; return *this; } bign operator / (const bign &b) { bign c, f = 0; for(int i = len-1; i >= 0; i--) { f = f*10; f.s[0] = s[i]; while(f >= b) { f -= b; c.s[i]++; } } c.len = len; c.clean(); return c; } bign operator /= (const bign &b) { *this = *this / b; return *this; } bign operator % (const bign &b) { bign r = *this / b; r = *this - r*b; return r; } bign operator %= (const bign &b) { *this = *this % b; return *this; } bool operator < (const bign &b) { 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) { 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) { return !(*this > b) && !(*this < b); } bool operator != (const bign &b) { return !(*this == b); } bool operator <= (const bign &b) { return *this < b || *this == b; } bool operator >= (const bign &b) { return *this > b || *this == b; } string str() const { string res = ""; for(int i = 0; i < len; i++) res = char(s[i]+'0') + res; return res; } }; 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; } int main() { bign a, b, c; cin>>a>>b>>c; a+=c; if(a>=b){ cout<<"little v"<<endl<<"height:"<<a<<endl; } else cout<<"little z"<<endl<<"height:"<<b<<endl; return 0; }