I (1 <= I <= 10^60)
by _SqrtTree_ @ 2022-08-14 10:19:36
爆long long
by _SqrtTree_ @ 2022-08-14 10:20:05
@[Dream_Crosyia](/user/663435) `long long` 必死。
考虑奇偶性，发现只需判断末尾数的奇偶性。
by JackMerryYoung @ 2022-08-14 10:23:53
# 100分代码
~~我也调了很久~~
```
#include<iostream>
#include<cstdio>
#include<stdio.h>
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll n;
string s;
int main()
{
cin>>n;
getchar();
for(ll i=1;i<=n;i++)
{
getline(cin,s);
if(s[s.size()-1]=='0'||s[s.size()-1]=='2'||s[s.size()-1]=='4'||s[s.size()-1]=='6'||s[s.size()-1]=='8')
cout<<"even"<<endl;
else
cout<<"odd"<<endl;
}
return 0;
}
```
## 高精度（极其简单）
by poor_OIer @ 2022-09-03 21:35:00
简单易懂的AC代码
```cpp
#include<bits/stdc++.h>
using namespace std;
int main(){
int n;
string sum;
cin>>n;
for(int i=1;i<=n;i++)
{
cin>>sum;
if((sum[sum.size()-1]-'0')&1) cout<<"odd\n";
else cout<<"even\n";
}
return 0;
}
```
# ~~高精是个好东西~~
by SaintRelief @ 2022-11-13 19:38:33