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CodeForces - 622A Infinite Sequence （思想）水

2016-03-12 14:03 211 查看
CodeForces
- 622A

Infinite Sequence

Time Limit: 1000MS Memory Limit: 262144KB 64bit IO Format: %I64d & %I64u
Submit Status

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 number 1 is
written out, then the numbers from 1 to 2, then the numbers from 1to 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 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


Source

Educational Codeforces Round 7

//题意：

给出如上的数列，输入一个n，问在这个数列上第n个位置上的数是啥。

//思路：直接模拟就行了，没啥技巧

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define INF 0x3f3f3f3f
#define IN __int64#define ll long long
#define N 10010#define M 1000000007
#define PI acos(-1.0)
using namespace std;
int main()
{
int i,j,k;
IN n;
while(scanf("%I64d",&n)!=EOF)
{
ll k=1;
while(n>k)
{
n-=k;
k++;
}
printf("%I64d\n",n);
}
return 0;
}


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