Codeforces 622A Infinite Sequence 【数学】

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).

Sample test(s)

input

3

output

2

input

5

output

2

input

10

output

4

input

55

output

10

input

56

output

1

题意：类似1 1 2 1 2 3 1 2 3 4 ....的序列，问你第i位数是多少。

水题，二分或者分类讨论都可以。

AC代码：

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <algorithm>#include <queue>#include <stack>#include <map>#include <set>#include <vector>#include <string>#define INF 1000000#define eps 1e-8#define MAXN (10000+10)#define MAXM (2000000+10)#define Ri(a) scanf("%d", &a)#define Rl(a) scanf("%lld", &a)#define Rf(a) scanf("%lf", &a)#define Rs(a) scanf("%s", a)#define Pi(a) printf("%d\n", (a))#define Pf(a) printf("%.2lf\n", (a))#define Pl(a) printf("%lld\n", (a))#define Ps(a) printf("%s\n", (a))#define W(a) while((a)--)#define CLR(a, b) memset(a, (b), sizeof(a))#define MOD 1000000007#define LL long long#define lson o<<1, l, mid#define rson o<<1|1, mid+1, r#define ll o<<1#define rr o<<1|1#define PI acos(-1.0)#pragma comment(linker, "/STACK:102400000,102400000")#define fi first#define se secondusing namespace std;typedef pair<int, int> pii;int main(){    LL n; Rl(n); n *= 2;    LL mid = sqrt(n);    LL l = mid-1, r = mid+1;    LL pos;    if(l * (l+1) >= n)        pos = l;    else if(mid * (mid+1) >= n)        pos = mid;    else        pos = r;    n /= 2; pos--;    LL yu = n - pos * (pos+1) / 2;    Pl(yu);    return 0;}