B. Draw!（简单模拟）

最新推荐文章于 2020-04-17 09:00:27 发布

原创最新推荐文章于 2020-04-17 09:00:27 发布 · 349 阅读

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本文解析了CodeForces平台上的一道题目B.Draw!，介绍了比赛得分变化的规律及最大平局次数的算法实现，通过C++代码展示了如何计算从比赛开始到结束可能出现的最大平局次数。

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题目链接：http://codeforces.com/problemset/problem/1131/B

B. Draw!

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You still have partial information about the score during the historic football match. You are given a set of pairs (ai,bi)(ai,bi), indicating that at some point during the match the score was "aiai: bibi". It is known that if the current score is «xx:yy», then after the goal it will change to "x+1x+1:yy" or "xx:y+1y+1". What is the largest number of times a draw could appear on the scoreboard?

The pairs "aiai:bibi" are given in chronological order (time increases), but you are given score only for some moments of time. The last pair corresponds to the end of the match.

Input

The first line contains a single integer nn (1≤n≤100001≤n≤10000) — the number of known moments in the match.

Each of the next nn lines contains integers aiai and bibi (0≤ai,bi≤1090≤ai,bi≤109), denoting the score of the match at that moment (that is, the number of goals by the first team and the number of goals by the second team).

All moments are given in chronological order, that is, sequences xixi and yjyj are non-decreasing. The last score denotes the final result of the match.

Output

Print the maximum number of moments of time, during which the score was a draw. The starting moment of the match (with a score 0:0) is also counted.

Examples

input

Copy

output

Copy

input

Copy

output

Copy

input

Copy

1
5 4

output

Copy

Note

In the example one of the possible score sequences leading to the maximum number of draws is as follows: 0:0, 1:0, 2:0, 2:1, 3:1, 3:2, 3:3, 3:4.

题解：实在没什么可说的。

#include<bits/stdc++.h>
using namespace std;
int n,a,b,x,y,ans;
int main()
{
    cin>>n;
    for(int i=1;i<=n;i++)
    {
        cin>>a>>b;
        ans+=max(0,min(a,b)-max(x,y)+(x!=y));
        x=a;y=b;
    }
    cout<<ans+1<<endl;
    return 0;
}