codeforces 1455 B题

codeforces 1455 B
2021-01-14 打卡题，随便在codeforces上找了一道

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

题目描述：

You are standing on the OX-axis at point 0 and you want to move to an integer point ????>0.

You can make several jumps. Suppose you’re currently at point ???? (???? may be negative) and jump for the ????-th time. You can:

either jump to the point ????+????
or jump to the point ????−1.
What is the minimum number of jumps you need to reach the point ?????

Input
The first line contains a single integer ???? (1≤????≤1000) — the number of test cases.

The first and only line of each test case contains the single integer ???? (1≤????≤106) — the destination point.

Output
For each test case, print the single integer — the minimum number of jumps to reach ????. It can be proved that we can reach any integer point ????.

Example：

inputCopy
5
1
2
3
4
5

outputCopy
1
3
2
3
4

Note：
In the first test case ????=1, so you need only one jump: the 1-st jump from 0 to 0+1=1.

In the second test case ????=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.

思路分析：
这个题，画出对应的树，从根结点开始往下依次计算，画到大概5-6层的时候，规律已经很明显了，建议自己去找一找。

代码实现：

#include <iostream>
#include <cstdio>
using namespace std;
int main() {
int t;
scanf("%d",&t);
while(t--) {
int n;
scanf("%d",&n);
int sum = 0;
int i;
for(i=1;i<=n;i++) {
sum += i;
if(sum >= n)
break;
}
if(n != sum-1)
printf("%dn",i);
else
printf("%dn",i+1);
}
return 0;
}

最后

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