洛谷P1957 | 鱼塘

题目：P1957
题目的主要难点在于判断运算式的总长度，由于式子格式固定为 a ？b = c，扣除固定的两个运算符号，问题进一步转化为求每个数的位数(如为负数则加上负号)

对于一个正数n，我们log10(n)+1就是他的位数，但是对于非正数，我们无法对其取对数，因此我们需要自定义一个函数

int mylog10(int x) {

	if (x > 0) return log10(x) + 1; 
	if (x == 0) return 1;
	if (x < 0) return log10(-x) + 2; //负号+1
}

总代码

#include<bits/stdc++.h>
using namespace std;
int mylog10(int x) {

	if (x > 0) return log10(x) + 1;
	if (x == 0) return 1;
	if (x < 0) return log10(-x) + 2;
}

int main()
{
	ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
	int n; cin >> n;
	string i2;
	for (int i = 0; i < n; i++)
	{
		string i1;
		cin >> i1;
		if (i1 <= "z" && i1 >= "a")
		{
			int a, b;
			cin >> a >> b;
			if (i1 == "a")
			{
				cout << a << "+" << b << "=" << a + b << "\n";
				cout << mylog10(a) + mylog10(b) + mylog10(a + b)+ 2 << "\n";
			}
			if (i1 == "b")
			{
				cout << a << "-" << b << "=" << a - b << "\n";
				cout << mylog10(a) + mylog10(b) + mylog10(a - b) + 2 << "\n";
			}
			if (i1 == "c")
			{
				cout << a << "*" << b << "=" << a * b << "\n";
				cout << mylog10(a)  + mylog10(b)  + mylog10(a * b)  + 2 << "\n";
			}
			i2 = i1;
		}
		else
		{
			int a, b; cin >> b;
			a = stoi(i1);
			if (i2 == "a")
			{
				cout << a << "+" << b << "=" << a + b << "\n";
				cout << mylog10(a)  + mylog10(b)  + mylog10(a + b)  + 2 << "\n";
			}
			if (i2 == "b")
			{
				cout << a << "-" << b << "=" << a - b << "\n";
				cout << mylog10(a) + mylog10(b) + mylog10(a - b) + 2 << "\n";
			}
			if (i2 == "c")
			{
				cout << a << "*" << b << "=" << a * b << "\n";
				cout << mylog10(a)  + mylog10(b)  + mylog10(a * b)  + 2 << "\n";
			}
		}
	}
	return 0;
}