1455B Jumps(思维、规律)
B. Jumps
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
You are standing on the OX
-axis at point 0 and you want to move to an integer point x>0
.
You can make several jumps. Suppose you’re currently at point y
(y may be negative) and jump for the k
-th time. You can:
either jump to the point y+k
or jump to the point y−1
.
What is the minimum number of jumps you need to reach the point x
?
Input
The first line contains a single integer t
(1≤t≤1000
) — the number of test cases.
The first and only line of each test case contains the single integer x
(1≤x≤106
) — the destination point.
Output
For each test case, print the single integer — the minimum number of jumps to reach x
. It can be proved that we can reach any integer point x.
思路：先假设一直相加：1+2+3+4+5+6+…+k=sum，
这时通过观察可以发现假如第一步我选择退后则总和为
-1+2+3+4…+k=sum-2,假设在第二步倒退，1-1+3+4+…+k=sum-3，依此类推，我们可以得出我们从一一直加直到>=n(题目的输入),如果刚好=n，答案为k,如果sum-1==n，答案为g+1，我们只有在最后一步倒退可得，步数+1，如果sum>n，答案为g，我们将之前某一步设为到退即可。
Code
#include<iostream>
#include<string>
#include<map>
#include<algorithm>
#include<memory.h>
#include<cmath>
#include<stack>
#define pii pair<int,int>
#define FAST ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
using namespace std;
typedef long long ll;
const int Max = 2e5 + 5;
int lst[Max];
int Mod = 1e9 + 7;
int main()
{
FAST;
int t;cin >> t;
while (t--)
{
int n;cin >> n;
int i = 0, g = 0;
while (i < n)
{
i += (++g);
}
if (i == n)cout << g << endl;
else if (i - 1 == n)cout << g + 1 << endl;
else cout << g << endl;
}
}
