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题目描述
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
输入
The input is given from Standard Input in the following format:
X
输出
Print the earliest possible time for the kangaroo to reach coordinate X.
样例输入 Copy
6
样例输出 Copy
3
提示
The kangaroo can reach his nest at time 3 by jumping to the right three times, which is the earliest possible time.
题意: 从原点出发,在时间i可以走i的路程,可以向左向右或不动。问到达x的最短时间。
思路:
(1)贪心:一直往右走是时间最短的。会有两种情况:在i时恰好到达x,或在i时超过x;对于前者一定是最短时间,对于后者我们可以在时间i之前选择往左走,从而使i时恰好到达x;这也就引出了方法二
(2)前缀和+二分:基于(1)的分析我们可以维护一个前缀和数组。找数组中第一个>=x的位置即可。
代码:
(1)贪心(模拟)
#include<bits/stdc++.h> using namespace std; int x,start; void solve(){ cin>>x; for(int i=1;i<=x;i++){ start+=i; if(start>=x){ cout<<i<<endl; break; } } } int main(){ solve(); return 0; }
(2)前缀和+二分(手写二分)
#include<bits/stdc++.h> using namespace std; const int maxn=50000; int a[maxn],x; void solve(){ for(int i=1;i<maxn;i++) a[i]=a[i-1]+i; cin>>x; int l=0,r=maxn-1,res=0; while(l<=r){ int mid=l+(r-l)/2; if(a[mid]>=x) r=mid-1; else if(a[mid]<x)l=mid+1; } cout<<l<<endl; } int cutele(){ solve(); return 0; }
(3)STL 二分
#include<bits/stdc++.h> using namespace std; const int maxn=50000; int a[maxn],x; void solve(){ for(int i=1;i<maxn;i++) a[i]=a[i-1]+i; cin>>x; int ans=lower_bound(a,a+maxn,x)-a; cout<<ans<<endl; } int cutele(){ solve(); return 0; }
知识补充:
1.如果用mid=(left+right)/2,如果right是int上限一半还多,那么可能导致溢出,使用mid=left+(right-left)/2可以代替以避免溢出
2.lower_bound(first,last,val)函数在[first,last)进行二分查找返回大于或等于val的第一个元素的位置,如果是数组,返回该位置指针,若没有则返回last的位置。upper_bound()与lower_bound()相似,查找的是第一个大于val的元素的位置。举个栗子0v0:数组a={1,2,4,4,4,5,7,9,9,10,13,17},想查找4第一次出现位置,lower_bound(a,a+n,4)-a,返回值是2.
参考资料: 二分总结_
算法复习——二分(二分查找,stl lower_bound()、upper_bound()木棒切割,快速幂)
保持热爱,奔赴山海。