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Time limit : 2sec / Memory limit : 256MB

Score : 200 points

Problem Statement

There is a kangaroo at coordinate 0 on an infinite number line that runs from left to right, at time 0. During the period between time i−1 and time i, the kangaroo can either stay at his position, or perform a jump of length exactly i to the left or to the right. That is, if his coordinate at time i−1 is x, he can be at coordinate x−i,x or x+i at time i. The kangaroo's nest is at coordinate X, and he wants to travel to coordinate X as fast as possible. Find the earliest possible time to reach coordinate X.

Constraints

• X is an integer.
• 1≤X≤109

題意就是：起點在數軸的0點上，第i個時間可以走長度i，可以向左也可以向右。爲要到達x位置，最少需要的時間。

思路：題解很簡單一句話：The answer is the minimum t such that 1 + 2 + ... + t ≥ X.比賽時我也是按照這個思路寫的，什麼意思呢？

如果前t分鐘一直向左走，加起來的路程如果恰好等於x，那一定就是時間t最短。

可如果超過了呢？超過的距離是d，那麼滿足        d < sum(t+1) - sum(t)，d就是[0, t]之間的數，0的情況說了沒問題。

不是0呢，不論多了幾，肯定是之前出現的某個距離，根據d的範圍就知道了，那就之前那個距離的時候不走，這多出

的距離不就可以減去了嗎。所以這樣答案也是成立的。這樣就對了。

#include <bits/stdc++.h>
const int N = 44721;
using namespace std;
int sum[N];
int main()
{
 
 
    for (int i = 1; i <= N; i++)
    {
        sum[i] = sum[i-1] + i;
    }
    int x;
    while (~scanf("%d", &x))
    {
        int loc = lower_bound(sum, sum + N, x) - sum;
  
        printf("%d\n", loc);
    }   
    return 0;
}