Educational Codeforces Round 7-A. Infinite Sequence(模拟)

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本文介绍了一个无限整数序列的构造方式及其求值方法。该序列从1开始，逐渐增加序列的最大值，直到包含所有正整数。文章提供了一种快速定位特定位置数值的算法，并附带AC代码实现。

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




思路：
   一开始打算用n*n+1/2来做，结果没想到竟然爆long long了，原来直接减就行了。一开始认为这个方法会超时才用公式，结果竟然不超时，看来又想错了时间了。




AC代码：


#include<iostream>
#include<algorithm>
#include<cstring>
#include<string>
#include<cstdio>
#include<cmath>
#include<ctime>
#include<cstdlib>
#include<queue>
#include<vector>
#include<set>
using namespace std;
const int T=10000100;
#define inf 0x3f3f3f3fL
#define mod 1000000000
typedef long long ll;
typedef unsigned long long ULL;

int main()
{
#ifdef zsc 
	freopen("input.txt","r",stdin); 
#endif
	ll n,m,i,j=0,k;
	while(~scanf("%I64d",&n))
	{
		for(i=1;i<n;++i)n-=i;
		printf("%I64d\n",n);
	}
    return 0;
}