codeforces_622A. Infinite Sequence

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

水题来着，用sqrt乱搞都过了（也类似于二分），看网上有两种思路，纯数学和二分，都学习一下吧。

#include <iostream>#include <cstdio>#include <cstdlib>#include <cmath>#include <cstring>using namespace std;//sqrt乱搞int main(){    long long n,ans;    scanf("%lld",&n);    long long s=(long long)sqrt(2*n);    long long sum=s*(s+1)/2;    if(sum>n)    {        ans=s-(sum-n);    }    else if(sum==n)    {        ans=s;    }    else if(sum<n)    {        ans=n-sum;    }    printf("%lld\n",ans);    return 0;}

#include<cstdio>  #include<cstring>  #include<iostream>  #include<algorithm>  using namespace std;  //数学int main()  {      long long n;      while(cin>>n)      {          long long temp;          for(long long i=1;n>0;i++)          {              temp=n;              n-=i;          }          cout<<temp<<endl;      }      return 0;  }  

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;//二分long long solve(long long n){    long long l=0,r=10000000,mid;    while(l<=r)    {        mid=(l+r)/2;        if(mid*(mid+1)/2>=n)r=mid-1;        else l=mid+1;    }    return l-1;}int main(){    long long n;    scanf("%lld",&n);    long long ans=solve(n);    printf("%lld\n",n-(ans)*(ans+1)/2);    return 0;}