題目連結：https://vjudge.net/contest/260134#problem/A
題目：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
2 11 3
0 1
1 1
Output
3
Input
0 5 2
Output
2
Input
1 3 2
1 2
Output
0
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.
題目大意：Vasya一共工作L分鐘，抽一根菸要a分鐘，有顧客時不能抽菸，求V他最多能抽幾根菸。
因為ti+li≤ti+1ti+li≤ti+1 所以只需要根據輸入順序計算Vasya有多少空閒時間進行抽菸。
AC程式碼：

#include<cstdio>
using namespace std;
int main(){
    int n,l,a;
    while(scanf("%d%d%d",&n,&l,&a)!=EOF){
        int temp=0,sum=0;
        int t,tt;
        for(int i=0;i<n;i++){
            scanf("%d%d",&t,&tt);
            sum+=(t-temp)/a;
            temp=t+tt;
        }
    sum+=(l-temp)/a;
    printf("%d\n",sum);
    return 0;
    }
}