The Frog‘s Games

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 5676    Accepted Submission(s): 2732

Problem Description

The
annual Games in frogs‘ kingdom started again. The most famous
game is the Ironfrog Triathlon. One test in the Ironfrog Triathlon
is jumping. This project requires the frog athletes to jump over the
river. The width of the river is L (1<= L <=
1000000000). There are n (0<= n <= 500000) stones lined up
in a straight line from one side to the other side of the river. The
frogs can only jump through the river, but they can land on the stones.
If they fall into the river, they
are out. The frogs was asked to
jump at most m (1<= m <= n+1) times. Now the frogs want to know
if they want to jump across the river, at least what ability
should they have. (That is the frog‘s longest jump distance).

Input

The input contains several cases. The first line of each case contains three positive integer L, n, and m.
Then
n lines follow. Each stands for the distance from the starting banks to
the nth stone, two stone appear in one place is impossible.

Output

For each case, output a integer standing for the frog‘s ability at least they should have.

Sample Input

6 1 2
2
25 3 3
11
2
18

Sample Output

4
11

Source

The 36th ACM/ICPC Asia Regional Dalian Site —— Online Contest

#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
const int INF= 0x3f3f3f3f ;
int d[550000];
int L;
int calc(int x)
{
    int now=0;
    int cnt=0;
    int ps=0;

    while(now< L)
    {
        now+= x;
        while(now >=d[ps+1]) ps++;
        now=d[ps];
        cnt++;
    }
    return cnt;
}
int main()
{
    int n, m;
    while(scanf("%d%d%d", &L, &n, &m) != EOF)
    {
        int minx =0 ;
        for(int i=1; i<= n; i++) scanf("%d", &d[i]);
        sort(d+1, d+1+n);
        d[0]= 0; d[n+1]= L;
        d[n+2]= INF ;
        for(int i=1; i<= n+1; i++)
            if(minx <d[i]- d[i-1])
                minx= d[i]- d[i-1];

        int left= minx, right= L;
        int res;
        while(left <= right){
            int mid= (left+right)>> 1;
            if(calc(mid) <= m)
            {
                res= mid;
                right= mid-1;
            }
            else left= mid+1;
        }
        printf("%d\n", res);
    }
    return 0;
}

hdoj 4004 The Frog's Games(二分)