CodeForces - 622A A. Infinite Sequence

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探讨一个特殊整数序列的生成规律，并提供一种高效算法来确定任意位置上的具体数值。此序列从1开始，逐渐扩展至更大的数列，如1, 1, 2, 1, 2, 3...直至无穷。文章详细解析了如何通过数学方法快速定位到序列中任一位置的数字。

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

解题思路

1,1,2,1,2,3,1,2,3,4,1,2,3,4,5…观察可得出规律1,2,3,4,5。。。第n项就是前面的和，所以用n减去sum，就是对应的数。由于数据太大，所以不能直接用for()换求和，求和公式s=（首项+末项）*项数/2;

AC代码：

#include<cstdio>
typedef long long ll;
int main(){
	ll n, sum = 0, mod;
	scanf("%lld", &n);
	for(ll i = 1; ; i++){
		sum = 0; 
		sum = ((1 + i) * i) / 2;
		if(sum > n) {
			sum = ((1 + i-1) * (i-1)) / 2;
			mod =  n - sum;
			if(mod == 0){
				printf("%lld", i-1);
				break;
			}
			else{
				printf("%lld", mod);
				break;
			}
		}
	}
	return 0;
}