CF1455 B. Jumps

連結 https://codeforces.com/contest/1455/problem/B

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
input
5
1
2
3
4
5
output
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.

題意
我們能對一個數進行兩種操作：
第一種是在第i次操作減一
第二種是在第i次操作加i
現在給你一個數，求最少操作次數能到這個數？

思路
思維題，我們看樣例發現
求在1+2+3+…+n哪個區間內，輸出i值即可，如果這個數是這個數列和減一的話，則我們要加一次操作，觀察可推出。
比如說
1 + 2 + 3 + 4 = 10
9的話只能1 + 2 + 3 + 4 - 1可得，其他區間內的數都能改變其中項為-1實現。

程式碼

#include <bits/stdc++.h>
typedef long long ll;
const ll mod = 1e9+7;
using namespace std;
namespace fastIO {
    inline ll qpow(ll a, ll b) {
        ll ans = 1, base = a;
        while (b) {
            if (b & 1) ans = (ans * base % mod +mod )%mod;
            base = (base * base % mod + mod)%mod;
            b >>= 1;
        }
        return ans;
    }
}
using namespace fastIO;
const int N = 1e6 + 5;
int Case,n;
int f[N]; 
int cnt;
void init(){
	for(int i=0;i<=N;i++){
		int t=(i*i+i)/2;
		if(t>N) break;
		f[cnt++]=t;
	}
}
int main(){
	Case=1;
	init();
	cout<<endl;
	scanf("%d",&Case);
	while(Case--){
		scanf("%d",&n);
		int vis;
		for(int i=1;i<=cnt;i++){
			if(f[i]>=n){
				vis = i;
				break;
			} 
		}
		if(f[vis]-1==n) printf("%d\n",vis+1);
		else printf("%d\n",vis);
	
 	}
	return 0;
}