1 solutions
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1
叛逆的代码(没人造数据的快乐)
#include <bits/stdc++.h> using namespace std; int n; long long ans = 0; int main () { scanf("%d",&n); for(int i = 1;i<=n;i++)ans+=i; printf("%lld",ans); return 0; }
更叛逆的代码(不是不让用公式吗???我偏用!!!哈哈哈哈……)
#include <bits/stdc++.h> using namespace std; int n; int main () { scanf("%d",&n); printf("%lld",(1+n)*n/2); return 0; }
- 1
Information
- ID
- 64
- Time
- 1000ms
- Memory
- 256MiB
- Difficulty
- 10
- Tags
- (None)
- # Submissions
- 7
- Accepted
- 3
- Uploaded By