CF#196DIV2:B-Xenia and Ringroad

发布时间:2017-6-26 11:38:49 编辑:www.fx114.net 分享查询网我要评论
本篇文章主要介绍了"CF#196DIV2:B-Xenia and Ringroad",主要涉及到CF#196DIV2:B-Xenia and Ringroad方面的内容,对于CF#196DIV2:B-Xenia and Ringroad感兴趣的同学可以参考一下。

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 ai 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 a1, a2, ..., am (1 ≤ ai ≤ 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. Sample test(s) 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. 题意:根据提示可知,这条路是个环形,之能顺时针走,问按照输入给出的顺序遍历所有点需要的时间   #include <stdio.h> #include <string.h> #include <algorithm> using namespace std; int main() { __int64 n,m,i,j,ans,a,b; while(~scanf("%I64d%I64d",&n,&m)) { ans = 0; b = 1; for(i = 1;i<=m;i++) { scanf("%I64d",&a); if(a>=b) ans = ans+a-b; else ans = ans+a+n-b; b = a; } printf("%I64d\n",ans); } return 0; }  

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