CodeForces-339B-Xenia and Ringroad （循环队列，水题）

标签：注意   initial   The   inpu   false   moved   思路   pac   循环队列   

Xenia lives in a city that has n houses built along the main ringroad. The ringroad houses are numbered 1 through n in the clockwise order. The ringroad traffic is one way and also is clockwise.

Xenia has recently moved into the ringroad house number 1. As a result, she‘s got m things to do. In order to complete the i-th task, she needs to be in the house number a i and complete all tasks with numbers less than i. Initially, Xenia is in the house number 1, find the minimum time she needs to complete all her tasks if moving from a house to a neighboring one along the ringroad takes one unit of time.

Input

The first line contains two integers n and m (2?≤?n?≤?105,?1?≤?m?≤?105). The second line contains m integers a 1,?a 2,?...,?a m (1?≤?a i?≤?n). Note that Xenia can have multiple consecutive tasks in one house.

Output

Print a single integer — the time Xenia needs to complete all tasks.

Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.

Examples

input

4 3
3 2 3

output

6

input

4 3
2 3 3

output

2

Note

In the first test example the sequence of Xenia‘s moves along the ringroad looks as follows: 1?→?2?→?3?→?4?→?1?→?2?→?3. This is optimal sequence. So, she needs 6 time units.

思路：

路径移动类似循环队列移动，注意点：使用long long 保证不溢出

#include<bits/stdc++.h>
using namespace std;
long long m, n, x = 1, s, y;
int main() {
	//freopen("in.txt","r",stdin);
	ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
	for (cin >> n >> m; m--; s += (n + y - x) % n, x = y)//x起点位置，y目标位置
		cin >> y;
	cout << s;
}

CodeForces-339B-Xenia and Ringroad （循环队列，水题）

标签：注意   initial   The   inpu   false   moved   思路   pac   循环队列   

原文地址：https://www.cnblogs.com/RioTian/p/13360617.html