大哥,这是高精度啊。。。QAQ
by LCGUO @ 2019-11-12 13:22:52
@[tt100723](/user/252401) 1 <= I <= 10^60
by k2saki @ 2019-11-12 13:23:20
@[tt100723](/user/252401) 用字符串!
by k2saki @ 2019-11-12 13:23:35
@[tt100723](/user/252401)
你一边输入一边测试就能把数组省掉
by JasonZRY @ 2019-11-12 13:25:50
像这样
```
#include<bits/stdc++.h>
using namespace std;
string s;
int n,a,l;
int main(){
scanf("%d",&n);
for(int i=1;i<=n;i++){
cin>>s;
l=s.size();
a=s[l-1]-'0';
if(a%2)printf("odd\n");
else printf("even\n");
}
}
```
by JasonZRY @ 2019-11-12 13:30:02
什么意思?我不懂。因为我才4年级。
题目没说要高精度。
by xxx听取AC声一片 @ 2019-11-13 07:44:14
各位大佬,是这样吗?
```cpp
#include<bits/stdc++.h>
using namespace std;
long long n,a;
int main()
{
cin>>n;
for(int i=1;i<=n;i++)
{
cin>>a;
if(a%2==0) cout<<"even"<<endl;
if(a%2==1) cout<<"odd"<<endl;
}
return 0;
}
```
by xxx听取AC声一片 @ 2019-11-13 07:48:08
@[tt100123](/user/252401) 不对,你的i是int型的,而n是long long型的,如果n大一点,你的i肯定加不到n那里去的。
而且这道题不能用long long(会炸),得用字符串。
```cpp
#include <iostream>
#include <string>
#define Y "even\n"
#define N "odd\n"
using namespace std;
string n;
string s;
int main() {
cin >> n;
while(cin >> s){
if((s[s.length()-1]-48)%2==0) cout << Y;
else cout << N;
}
return 0;
}
```
by Return_ @ 2020-01-27 17:09:45
@[v果宝v](/user/187259) 不需要用高精度
by Return_ @ 2020-01-27 17:10:10
对,没有必要
```cpp
#include<iostream>
using namespace std;
int main()
{
int a;
string n;
cin >> a;
while(cin >> n)
{
if(n[n.size() - 1] % 2 == 1)
{
cout << "odd" << endl;
}
else
{
cout << "even" << endl;
}
}
return 0;
}
```
by chenyitian @ 2021-03-04 22:34:54