Funky Numbers CodeForces - 192A

最新推荐文章于 2022-03-24 22:46:23 发布

转载最新推荐文章于 2022-03-24 22:46:23 发布 · 45 阅读

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原文链接：http://www.cnblogs.com/zgglj-com/p/6752116.html

本文介绍了一个有趣的问题：如何判断一个给定的整数是否可以表示为两个三角形数之和。通过具体的示例和C++代码实现，展示了寻找解决方案的过程。

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As you very well know, this year's funkiest numbers are so called triangular numbers (that is, integers that are representable as $\text{[math]}$ , where k is some positive integer), and the coolest numbers are those that are representable as a sum of two triangular numbers.

A well-known hipster Andrew adores everything funky and cool but unfortunately, he isn't good at maths. Given number n, help him define whether this number can be represented by a sum of two triangular numbers (not necessarily different)!

Input

The first input line contains an integer n (1 ≤ n ≤ 10⁹).

Output

Print "YES" (without the quotes), if n can be represented as a sum of two triangular numbers, otherwise print "NO" (without the quotes).

Example

Input

Output

YES

Input

Output

NO

Note

In the first sample number $\text{[math]}$ .

In the second sample number 512 can not be represented as a sum of two triangular numbers.

写得有点丑~~~

 1 #include<iostream>
 2 #include<algorithm>
 3 #include<cstdio>
 4 #include<math.h>
 5 #include<string>
 6 using namespace std;
 7 typedef long long ll;
 8 
 9 int n;
10 ll cal(ll ans){
11    return ans*(ans+1)/2;
12 }
13 
14 int main()
15 {  while(~scanf("%d",&n)){
16        int flag=0;
17        for(int i=1;i<=sqrt(2*n)+1;i++){
18               ll sum=n-cal(i);
19               ll l=i,r=sqrt(2*n)+1;
20               ll mid;
21               while(l<r){
22                     mid=(l+r)/2;
23                     if(cal(mid)==sum){
24                          flag=1;
25                          break;
26                     }
27                     if(cal(mid)<sum) l=mid;
28                     else r=mid;
29                     if(r-l==1){
30                          if(cal(r)==sum||cal(l)==sum){
31                              flag=1;
32                              break;
33                          }
34                          else break;
35                     }
36               }
37               if(flag) break;
38        }
39        if(flag) printf("YES\n");
40        else printf("NO");
41    }
42 }

转载于:https://www.cnblogs.com/zgglj-com/p/6752116.html