B. Jumps(模拟)

B. Jumps
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard 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.

Example
inputCopy
5
1
2
3
4
5
outputCopy
1
3
2
3
4
Note
In the first test case x=1, so you need only one jump: the 1-st jump from 0 to 0+1=1.

In the second test case x=2. You need at least three jumps:

the 1-st jump from 0 to 0+1=1;
the 2-nd jump from 1 to 1+2=3;
the 3-rd jump from 3 to 3−1=2;
Two jumps are not enough because these are the only possible variants:

the 1-st jump as −1 and the 2-nd one as −1 — you’ll reach 0−1−1=−2;
the 1-st jump as −1 and the 2-nd one as +2 — you’ll reach 0−1+2=1;
the 1-st jump as +1 and the 2-nd one as −1 — you’ll reach 0+1−1=0;
the 1-st jump as +1 and the 2-nd one as +2 — you’ll reach 0+1+2=3;
In the third test case, you need two jumps: the 1-st one as +1 and the 2-nd one as +2, so 0+1+2=3.

In the fourth test case, you need three jumps: the 1-st one as −1, the 2-nd one as +2 and the 3-rd one as +3, so 0−1+2+3=4.

题意：
从$0$点开始走
第$k$步你可以有两种选择：

1. 向前走$k$步　+k
2. 向后走$1$步　-1
   问你走到$n$时最少走多少步？

题解：
其实很简单的模拟，假设每次都往前走那么就是$1+2+3+4.....+x=y$就是$y=(n+1)\ast n/2$。
那么我们要走到 y-2我们就可以将$+1$改成$-1$。
走到$y-3$我们就可以将$+2$改成$-1$
…
走到$y-x+1$我们就可以将$+(x-2)$改成$-1$
其实就是在$y$上减去变化的步数$z$并且把他改成$-1$还要减$1$，所以我们最小就要减去$2$。对于$y-1$我们只能让他在到$y$后退一步了。

所以我们就推出，
(1). 如果$(n+1)\ast n/2=y$　那么就是走$n$步
(2). 如果$(n+1)\ast n/2=y+1$那么就是走$n$步再往后退一格　也就是$n+1$步。
(3). 其他其实就是跟我们分析的一样　走$n$步。

#include <bits/stdc++.h>
using namespace std;
#define ll long long
ll n;
ll sum;
ll dp[1000100];
string str;
int main() {
    
    
    int t;
    cin>>t;

    while(t--) {
    
    
        cin>>n;
        int i = 1;
        for(; i * (i + 1) / 2 < n; i ++);
        sum = i * (i + 1) / 2;
        if(sum == n) {
    
    
            cout << i << endl;
        } else if(sum == n + 1) {
    
    
            cout << i + 1 << endl;
        } else cout << i << endl;
    }
    return 0;
}