（PAT乙级刷题） 舍入

题目：

题解：

#include<iostream>
#include<string>
using namespace std;
int main(){
	int n, d, com, cnt, minus, all, carry, point, p, need, have, i, t;
	string s;
	cin >> n >> d;
	while(n--){
		cnt = minus = all = carry = point = p = need = have = 0;
		cin >> com >> s;
		if(s[0] == '-'){
			minus = 1;
			s.erase(0, 1);
		}
		while(p < s.size() && s[p] != '.') p++;//找到小数点位置
		if(p < s.size() && com != 2){//处理进位
			if(com == 1 && p + d + 1 < s.size() && s[p + d + 1] > '4') need = 1;
			else if(com == 3 && p + d + 1 < s.size() && s[p + d + 1] > '4'){
				if(s[p + d + 1] > '5') need = 1;
				else{
					for(int i = p + d + 2; i < s.size() && !have; i++){
						if(s[i] >= '1' && s[i] <= '9') have = 1;
					}
					if(have || (s[p + d] - '0') % 2 == 1) need = 1;
				}
			}
			carry = need;
			for(int i = p + d; i >= 0 && need; i--){//高精度加法
				if(s[i] == '.') continue;
				t = s[i] - '0' + carry;
				s[i] = t % 10 + '0';
				carry = t / 10;
			}
			if(carry) s = "1" + s;//最高位仍有进位时，数字前加一
		}
		for(int i = 0; i < s.size() && i < p + d && !all; i++){//是否全0
			if(s[i] >= '1' && s[i] <= '9') all = 1;
		}
		if(all && minus) cout << "-";//不是全0且有负号，才输出负号
		for(i = 0; i < s.size() && i <= p + d; i++){
			cout << s[i];
		}
		if(p == s.size()) {//整数时，添加小数点（D是正整数，所以一定需要输出小数点）
			cout << ".";
			p--;
		}
		while(i - p <= d + carry){//不足D位则补0
			cout << "0";
			i++;
		}
		cout << endl;
	}
	return 0;
}