阿凡提和巴依老爷

rc247chenbosheng1 LV 6 @ 2024-3-14 20:41:19

#include<bits/stdc++.h> 
using namespace std; 
int main(){ 
    int a,b,c; 
    cin>>a; 
    b=2; 
    c=pow(b,a); 
    cout<<c; 
    return 0; }

2
rc247shezhiyi1 LV 5 @ 2024-4-21 10:28:01

#include using namespace std; int main() { int i,n,m=1,sum=0; cin>>n; for(i=1;i<=n;i++) { m*=i; sum+=m; } cout<<sum<<endl; return 0; }
1
taotao1 LV 6 @ 2024-7-15 20:27:20
题目分析：

由题意，可以分析得出第 $n$ 个格子中应放置 $2^n$ 粒麦子，故使用左移运算符 << 来求 $2$ 的 $n$ 次幂即可。

代码实现：
```
#include<cstdio>
int main(){
	int n;
	scanf("%d",&n);
	printf("%d",1<<n);
	return 0;
}
```
记得给五星好评哦喵~ 谢谢客官啦~
0
rc247suzihan1 LV 4 @ 2024-4-14 9:44:56

#include<bits/stdc++.h> using namespace std; int main(){ int a,b,c; cin>>a; b=2; c=pow(b,a); cout<<c; return 0; }
-1
rc245liai1 LV 1 @ 2024-3-31 15:43:39
#include using namespace std; int main(){ long long m=1,n; cin>>n; for (int i=1;i<=n ;i++ ) { m=m*2; } cout<<m; return 0;
```
}
```

-1

ws244yangyuchen1 LV 2 @ 2024-3-31 10:17:27

#include<bits/stdc++.h> 
using namespace std; 
int main(){ 
    int a,b,c; 
    cin>>a; 
    b=2; 
    c=pow(b,a); 
    cout<<c; 
    return 0; }

-3
rc245sunruoyang1 LV 5 @ 2024-3-31 15:36:26

#include<bits/stdc++.h> using namespace std; int main(){ int a,b,c; cin>>a; b=2; c=pow(b,a); cout<<c; return 0; }
-3
ws247liangkun1 LV 1 @ 2024-3-29 17:38:40

#include<bits/stdc++.h> using namespace std; int main(){ long long n,sun=0,cnt =1; cin>>n; for(int i = 1;i<=n;i++){ cnt = 2*cnt; } cout<<cnt; return 0; }
-3
ws247gaojian1 LV 5 @ 2024-3-29 17:05:43
1. #include<bits/stdc++.h> using namespace std; int main(){ long long n,sum=0,cnt =1; cin>>n; for(int i = 1;i<=n;i++){ cnt = 2*cnt; } cout<<cnt; return 0;
2. }
-4
rc245wangchenran1 LV 5 @ 2024-3-31 10:31:13

#include<bits/stdc++.h> using namespace std; int main(){ long long n,m=1,sum=0; cin>>n; for(int i=1;i<=n;i++){ m=2*m; } cout<<m; return 0; }
-4
rc246lihao1 LV 6 @ 2024-2-23 21:46:19

#include<bits/stdc++.h> using namespace std; int main(){ int a,b; cin>>a; b=pow(2,a); cout<<b; return 0; }