CodeForces round 753 problem B Odd Grasshopper（奇怪的蚱蜢）

题目翻译：

The grasshopper(蚂蚱) is located on the numeric axis （数轴）at the point with coordinate（坐标） x0.

一个蚱蜢在数轴上，初始坐标为x0

Having nothing else to do he starts jumping between integer points on the axis. Making a jump from a point with coordinate x with a distance d to the left moves the grasshopper to a point with a coordinate x−d, while jumping to the right moves him to a point with a coordinate x+d.

The grasshopper is very fond of positive integers（正整数）, so for each integer i starting with 1 the following holds: exactly iiminutes after the start he makes a jump with a distance of exactly i. So, in the first minutes he jumps by 1, then by 2, and so on.

开始后一分钟移动距离为1，两分钟移动距离为2，.....,

The direction of a jump is determined as follows: if the point where the grasshopper was before the jump has an even（偶数的） coordinate, the grasshopper jumps to the left, otherwise he jumps to the right.

移动方向由以下规则决定：

如果坐标是偶数，向左移

如果坐标是奇数，向右移

For example, if after 18 consecutive jumps he arrives at the point with a coordinate 77, he will jump by a distance of 19 to the right, since 77 is an odd number, and will end up at a point 7+19=267+19=26. Since 26 is an even number, the next jump the grasshopper will make to the left by a distance of 20, and it will move him to the point 26−20=626−20=6.

Find exactly which point the grasshopper will be at after exactly n jumps.

问题：求跳n次后，蚱蜢的坐标

Input

The first line of input contains an integer tt(1≤t≤10^4) — the number of test cases.

Each of the following tt lines contains two integers x0

and n — the coordinate of the grasshopper's initial position and the number of jumps.

输入：

第一行输入case个数

以下各行输入两个数，第一个数是起始坐标，第二个数是跳n次后的坐标

Output

Print exactly tt lines. On the ii-th line print one integer — the answer to the ii-th test case — the coordinate of the point the grasshopper will be at after making nn jumps from the point x0.

输出：

每一行表示每一个case jump 结束后的坐标

完整问题：

一个蚱蜢在数轴上，初始坐标为x0，开始后一分钟移动距离为1，两分钟移动距离为2，.....,

移动方向由以下规则决定：如果坐标是偶数，向左移如果坐标是奇数，向右移

问题：求跳n次后，蚱蜢的坐标

输入：

第一行输入case个数

以下各行输入两个数，第一个数是起始坐标，第二个数是跳n次后的坐标

输出：

每一行表示每一个case jump 结束后的坐标

思路：

奇数+奇数=偶数

奇数+偶数=奇数

偶数+偶数=偶数

一步步写可以看出规律

4个一组和为0，xo分奇偶讨论，n分模4讨论

代码：

#include<iostream>
#include<vector>using namespace std;long long jump(long long x, long long n)
{if (x % 2 != 0){if (n % 4 == 0){return x;}else if (n % 4 == 1){return x + n;}else if (n % 4 == 2){return x - 1;}else if (n % 4 == 3){return x - n - 1;}}else{if (n % 4 == 0){return x;}else if (n % 4 == 1){return x - n;}else if (n % 4 == 2){return x + 1;}else if (n % 4 == 3){return x + n + 1;}}
}
int main()
{long long m;cin >> m;vector<vector< long long>>input;vector< long long>temp;long long x1, n1;while (m--){cin >> x1;cin >> n1;temp.push_back(x1);temp.push_back(n1);input.push_back(temp);temp.clear();}for (int i = 0; i < input.size(); i++){x1 = input[i][0];n1 = input[i][1];cout << jump(x1, n1) << endl;}
}