CF 622 A. Infinite Sequence【找无限数列中第i个数是几】

A. Infinite Sequence

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

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).

Examples

input

3

output

2

input

5

output

2

input

10

output

4

input

55

output

10

input

56

output

1

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <queue>
#include <stack>
#include <algorithm>
#define Wi(a) while((a)--)
#define Si(a) scanf("%d", &a)
#define Sl(a) scanf("%I64d", &a)
#define Pi(a) printf("%d\n", (a))
#define Pl(a) printf("%I64d\n", (a))
#define CLR(a, b) memset(a, (b), sizeof(a))
#define INF 1e9
#define LL __int64using namespace std;

int main()
{
LL n;
while(Sl(n)==1)
{
LL  l, r, mid;
l = 0, r = INF;
while(r-l > 1)
{
mid = (l+r)>>1;
if(mid*(mid+1)>>1 <= n)  l = mid;
else  r = mid;
}
if((l+1)*l>>1 == n)  Pl(l);
else
{
r = (1+l)*l>>1;
Pl(n-r);
}
}
return 0;
}