Infinite Sequence

Infinite Sequence

Crawling in process...Crawling failedTime Limit:1000MS    Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

SubmitStatus Practice CodeForces 622A

Description

Consider the infinite sequence of integers: 1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5.... The sequence is built in the following way: at first the number1 is written out, then the numbers from 1 to 2, then the numbers from 1 to 3, then the numbers from 1 to 4 and so on. Note that the sequence contains numbers, not digits. For example number10 first appears in the sequence in position 55 (the elements are numerated from one).

Find the number on the n-th position of the sequence.

Input

The only line contains integer n (1 ≤ n ≤ 10¹⁴) — the position of the number to find.

Note that the given number is too large, so you should use 64-bit integer type to store it. In C++ you can use thelong long integer type and in Java you can use long integer type.

Output

Print the element in the n-th position of the sequence (the elements are numerated from one).

Sample Input

Input

3

Output

2

Input

5

Output

2

Input

10

Output

4

Input

55

Output

10

Input

56

Output

1

1+2……+m=m*(m+1)/2<n来确定m的大小。n (1 ≤ n ≤ 10¹⁴) ，通过for循环来查找m的值太费时，太麻烦。经过化简m*(m+1)<2*n来确定m，m近似等于k=sqrt(2*n)。可以看做是近似直接定位，减少时间。在取k值时要仔细分析.考虑一些特殊情况。如当n=2时，k=2,由于第二个位置是1，所以m=k-1=1，当n=3时，m=k-1=1，第3个位置应是2

My solution:

/*2016.3.12*/

#include<stdio.h>#include<math.h>#include<algorithm>using namespace std;int main(){long long n,m,ans,k,k1,k2,k3;int i,j;while(scanf("%I64d",&n)==1){k=sqrt(2*n);k=k-1;k1=(k+1)*k/2;k2=(k+1)*(k+2)/2;k3=(k+2)*(k+3)/2;if(n>k1&&n<=k2)m=n-k1;elseif(n>k2&&n<=k3)m=n-k2;printf("%I64d\n",m);}return 0;}