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.
Entry
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.
Export
Line 1: A single integer that is the maximum of the shortest distance a cow has to jump after removing M rocks
Sample input
25 5 2
2
14
11
21
17
Sample Output
4
prompt
. 1 #include <the iostream> 2 #include <CString> . 3 #include <algorithm> . 4 the using namespace STD; . 5 typedef Long Long LL; . 6 LL L, N, M; . 7 LL ARR [ 50005 ]; . 8 . 9 BOOL Check (LL MID) { /// half a shortest distance and the number of comparison to the actual throw to throw number 10 LL NUM = 0 , SUM = 0 ; . 11 for ( int I = . 1 ; I <= N + . 1 ; I ++ ) { 12 ARR = + [I] SUM - ARR [I- . 1 ]; 13 is // IF (SUM <= MID) NUM ++; // at this distance to remove the 14 IF (SUM <MID) NUM ++ ; 15 /// such as peracetic this stone meet others put off after his look before it is in line with the sum summing 16 the else { . 17 sum = 0 ; 18 is } . 19 } 20 is // return NUM <= M; 21 is return NUM> M; 22 is } 23 is 24 LL Binary (LL left, right LL) { 25 LL ANS; /// shortest distance is the removal of 26 while(left<=right){ 27 ll mid=left+right>>1; 28 if(check(mid)) right=mid-1; ///移除多了 29 else left=mid+1,ans=mid; 30 } 31 return ans; 32 } 33 34 int main(){ 35 ios::sync_with_stdio(false); 36 cin>>L>>N>>M; 37 ll maxx=0; 38 arr[0]=0 ; 39 for ( int I = . 1 ; I <= N; I ++) CIN >> ARR [I]; 40 ARR [N + . 1 ] = L; 41 is Sort (ARR + . 1 , ARR + N + . 1 + . 1 ); 42 is COUT Binary << ( . 1 , ARR [N + . 1 ]) << endl; // the shortest distance from the beginning 43 is return 0 ; 44 is }