标签:ons `` order item roc star uri nts for
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.
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.
Line 1: A single integer that is the maximum of the shortest distance a cow has to jump after removing M rocks
25 5 2
2
14
11
21
17
4
Before removing any rocks, the shortest jump was a jump of 2 from 0 (the start) to 2. After removing the rocks at 2 and 14, the shortest required jump is a jump of 4 (from 17 to 21 or from 21 to 25).
用二分,找到是否符合条件的间距
There is a literal backtick (`) here.
```#include
typedef long long ll;
typedef long double ld;
const ll MOD=1e9+7;
using namespace std;
ll a[50005];
ll l,n,num;
bool judge(ll mid)
{
int bits=0;
for(int i=0;i<n;i++)
{
int ans=0;
for(int j=1;j<=n-i;j++)
{
if(a[i+j]-a[i]<mid)
{
ans++;
bits++;
}else
break;
}
i+=ans;
if(bits>num)
{
return true;
}
}
return false;
}
int main()
{
a[0]=0;
cin>>l>>n>>num;
for(int i=1;i<=n;i++)
cin>>a[i];
a[n+1]=l;
n++;
sort(a,a+n+1);
ll left=0,right=l,mid=(left+right)/2;
while(right-left>1)
{
if(judge(mid))
{
right=mid;
}else
{
left=mid;
}
mid=(left+right)/2;
}
while(!judge(mid))mid++;
cout<<mid-1;
} ```
标签:ons `` order item roc star uri nts for
原文地址:https://www.cnblogs.com/wzl19981116/p/9347693.html