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
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| 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
}
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总代码
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| #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;
}
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