Educational Codeforces Round 7 A. Infinite Sequence

最新推荐文章于 2023-12-05 21:21:53 发布

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本文介绍了一个无限整数序列的构造方式，并提供了一种算法来找到该序列中任意位置的数值。通过对序列特性的分析，文章给出了解决方案的代码实现。

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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 number 1 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 number 10 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 the long 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 test(s)

Input

Output

Input

Output

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1



题意：按照题目给的序列，求第n个数字是多少

思路：水题，直接写就好





ac代码：

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stack>
#include<set>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#define MAXN 1010000
#define LL long long
#define ll __int64
#define INF 0xfffffff
#define mem(x) memset(x,0,sizeof(x))
#define PI acos(-1)
using namespace std;
int gcd(int a,int b){return b?gcd(b,a%b):a;}
LL powmod(LL a,LL b,LL MOD){LL ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
//head
int main()
{
	ll n;
	while(scanf("%I64d",&n)!=EOF)
	{
		ll k=1;
		while(n-k>0)
		{
			n-=k;
			k++;
		}
		printf("%I64d\n",n);
	}
	return 0;
}