Codeforces Round #514 (Div. 2).A. Cashier

最新推荐文章于 2018-10-20 19:20:25 发布

原创最新推荐文章于 2018-10-20 19:20:25 发布 · 295 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#Codeforces Round #514 (Div. 2)

Codeforces 专栏收录该内容

15 篇文章

订阅专栏

探讨在服务固定客户群的同时，收银员如何安排休息时间以达到最大休息次数的算法问题。输入包括工作时间长度、客户到达时间和服务时间，输出为收银员在不耽误工作的前提下能采取的最大休息次数。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

A. Cashier

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Vasya has recently got a job as a cashier at a local store. His day at work is LL minutes long. Vasya has already memorized nn regular customers, the ii-th of which comes after titi minutes after the beginning of the day, and his service consumes lili minutes. It is guaranteed that no customer will arrive while Vasya is servicing another customer.

Vasya is a bit lazy, so he likes taking smoke breaks for aa minutes each. Those breaks may go one after another, but Vasya must be present at work during all the time periods he must serve regular customers, otherwise one of them may alert his boss. What is the maximum number of breaks Vasya can take during the day?

Input

The first line contains three integers nn, LL and aa (0≤n≤1050≤n≤105, 1≤L≤1091≤L≤109, 1≤a≤L1≤a≤L).

The ii-th of the next nn lines contains two integers titi and lili (0≤ti≤L−10≤ti≤L−1, 1≤li≤L1≤li≤L). It is guaranteed that ti+li≤ti+1ti+li≤ti+1 and tn+ln≤Ltn+ln≤L.

Output

Output one integer — the maximum number of breaks.

Examples

input

Copy

2 11 3
0 1
1 1

output

input

0 5 2

output

input

1 3 2
1 2

output

Note

In the first sample Vasya can take 33 breaks starting after 22, 55 and 88 minutes after the beginning of the day.

In the second sample Vasya can take 22 breaks starting after 00 and 22 minutes after the beginning of the day.

In the third sample Vasya can't take any breaks.

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define M(a) memset(a,0,sizeof(a))
#define rep(i,a) for(int i=1;i<=a;i++)

struct node{
    int arr;
    int las;
}b[100010];


bool cmp(node a,node b){
    return a.arr<b.arr;
}

int main()
{
    int n,m,L,a;
    while(cin>>n>>L>>a)
    {
        memset(b,0,sizeof(b));
        for(int i=0;i<n;i++){
            cin>>b[i].arr;
            cin>>b[i].las;
        }
        sort(b,b+n,cmp);
        int ans=0;
        ans+=b[0].arr/a;
        for(int i=1;i<n;i++){
            ans+=((b[i].arr-(b[i-1].las+b[i-1].arr))/a);
        }
        ans+=((L-(b[n-1].arr+b[n-1].las))/a);
        cout<<ans<<endl;
    }
    return 0;
}