POJ

Farmer John has a problem: the dirt road from his farm to town has suffered in the recent rainstorms and now contains (1 <= N <= 10,000) mud pools. 

Farmer John has a collection of wooden planks of length L that he can use to bridge these mud pools. He can overlap planks and the ends do not need to be anchored on the ground. However, he must cover each pool completely. 

Given the mud pools, help FJ figure out the minimum number of planks he needs in order to completely cover all the mud pools.

Input

* Line 1: Two space-separated integers: N and L 

* Lines 2..N+1: Line i+1 contains two space-separated integers: s_i and e_i (0 <= s_i < e_i <= 1,000,000,000) that specify the start and end points of a mud pool along the road. The mud pools will not overlap. These numbers specify points, so a mud pool from 35 to 39 can be covered by a single board of length 4. Mud pools at (3,6) and (6,9) are not considered to overlap. 

Output

* Line 1: The miminum number of planks FJ needs to use.

Sample Input

3 31 613 178 12

Sample Output

5

Hint

INPUT DETAILS: 

FJ needs to use planks of length 3 to cover 3 mud pools. The mud pools cover regions 1 to 6, 8 to 12, and 13 to 17. 

OUTPUT DETAILS: 

FJ can cover the mud pools with five planks of length 3 in the following way: 

                   111222..333444555....                   .MMMMM..MMMM.MMMM....                   012345678901234567890

对泥潭按起点排序后，定义迭代变量len记录当前走到哪里，走完所有泥潭自然得到结果，注意一个泥潭的终点是p[i].e-1

代码：

#include<iostream>#include<string>#include<cstdio>#include<algorithm>#include<cmath>#include<iomanip>#include<queue>#include<cstring>#include<map>#include<vector>using namespace std;typedef long long ll;#define M 102int n,l;struct node{    ll s,e;    bool operator <(const node & obj)const    {        return s<obj.s;    }}p[10005];int main(){    int i,temp;    scanf("%d%d",&n,&l);    ll len,num;    for(i=0;i<n;i++)        scanf("%I64d%I64d",&p[i].s,&p[i].e);    sort(p,p+n);    num=0; len=p[0].s-1;    for(i=0;i<n;i++)    {        if(len<p[i].s) len=p[i].s-1;        while(len<p[i].e-1)        {            if((p[i].e-1-len)>l) {temp=(p[i].e-1-len)/l; num+=temp; len+=l*temp;}  //这一步对节省时间很重要            else { num++; len+=l;}        }    }    printf("%I64d",num);    return 0;}