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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

题意就是：起点在数轴的0点上，第i个时间可以走长度i，可以向左也可以向右。为要到达x位置，最少需要的时间。

思路：题解很简单一句话：The answer is the minimum t such that 1 + 2 + ... + t ≥ X.比赛时我也是按照这个思路写的，什么意思呢？

如果前t分钟一直向左走，加起来的路程如果恰好等于x，那一定就是时间t最短。

可如果超过了呢？超过的距离是d，那么满足        d < sum(t+1) - sum(t)，d就是[0, t]之间的数，0的情况说了没问题。

不是0呢，不论多了几，肯定是之前出现的某个距离，根据d的范围就知道了，那就之前那个距离的时候不走，这多出

的距离不就可以减去了吗。所以这样答案也是成立的。这样就对了。

#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;}