Description

Every year the cows hold an event featuring a peculiar version of hopscotch that involves carefully jumping from rock to rock in a river. The excitement takes place on a long, straight river with a rock at the start and another rock at the end, L units away from the start (1 <= L <= 1,000,000,000). Along the river between the starting and ending rocks, N (0 <= N <= 50,000) more rocks appear, each at an integral distance Di from the start (0 < Di < L). To play the game, each cow in turn starts at the starting rock and tries to reach the finish at the ending rock, jumping only from rock to rock. Of course, less agile cows never make it to the final rock, ending up instead in the river. Farmer John is proud of his cows and watches this event each year. But as time goes by, he tires of watching the timid cows of the other farmers limp across the short distances between rocks placed too closely together. He plans to remove several rocks in order to increase the shortest distance a cow will have to jump to reach the end. He knows he cannot remove the starting and ending rocks, but he calculates that he has enough resources to remove up to M rocks (0 <= M <= N). FJ wants to know exactly how much he can increase the shortest distance *before* he starts removing the rocks. Help Farmer John determine the greatest possible shortest distance a cow has to jump after removing the optimal set of M rocks.

數軸上有n個石子，第i個石頭的坐標為Di，現在要從0跳到L，每次條都從一個石子跳到相鄰的下一個石子。現在FJ允許你移走M個石子，問移走這M個石子後，相鄰兩個石子距離的最小值的最大值是多少。

Input

* Line 1: Three space-separated integers: L, N, and M * Lines 2..N+1: Each line contains a single integer indicating how far some rock is away from the starting rock. No two rocks share the same position.

Output

* Line 1: A single integer that is the maximum of the shortest distance a cow has to jump after removing M rocks

Sample Input

Sample Output

#include <cmath>
#include 
 
<cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
char buf[10000000], *ptr = buf - 1;
inline int readint(){
    int f = 1, n = 0;
    char ch = *++ptr;
    while(ch < ‘0‘ || ch > ‘9‘){
        if(ch == ‘-‘) f = -1;
        ch = *++ptr;
    }
    while(ch <= ‘9‘ && ch >= ‘0‘){
        n = (n << 1) + (n << 3) + ch - ‘0‘;
        ch = *++ptr; 
    }
    return f * n;
}
const int maxn = 500000 + 10;
int L, N, M, d[maxn];
inline bool Judge(int mid){
    int tot = 0, last = 1;
    for(int i = 2; i <= N; i++){
        if(d[i] - d[last] < mid) tot++;
        else last = i;
    }
    return tot <= M;
}
int main(){
    fread(buf, sizeof(char), sizeof(buf), stdin);
    L = readint();
    N = readint() + 2;
    M = readint();
    for(int i = 2; i < N; i++) d[i] = readint();
    d[1] = 0;
    d[N] = L;
    sort(d + 1, d + N + 1);
    int l = 1, r = L, mid, ans;
    while(l <= r){
        mid = l + r >> 1;
        if(Judge(mid)){
            ans = mid;
            l = mid + 1;
        }
        else r = mid - 1;
    }
    printf("%d\n", ans);
    return 0;
}

[BZOJ1650][Usaco2006 Dec]River Hopscotch 跳石子