思路：依次把x【】数组的元素加入set，再二分出当前入set元素在set中的上界r和下界l,则这段区间能放的最多的战舰数量就由原来的（r-l)/(a+1)减少为（x[i]-l）/(a+1)+(r-x[i])/(a+1),
每次判断加入新元素后能放最多战舰数sum与k的大小即可。
1 #include <iostream> 2 #include <queue> 3 #include <stack> 4 #include <cstdio> 5 #include <vector> 6 #include <map> 7 #include <set> 8 #include <bitset> 9 #include <algorithm> 10 #include <cmath> 11 #include <cstring> 12 #include <cstdlib> 13 #include <string> 14 #include <sstream> 15 #include <time.h> 16 #define x first 17 #define y second 18 #define pb push_back 19 #define mp make_pair 20 #define lson l,m,rt*2 21 #define rson m+1,r,rt*2+1 22 #define mt(A,B) memset(A,B,sizeof(A)) 23 #define mod 1000000007 24 using namespace std; 25 typedef long long LL; 26 const int N=2e5+10; 27 const int inf = 0x3f3f3f3f; 28 const LL INF=0x3f3f3f3f3f3f3f3fLL; 29 int x[N]; 30 set<int> Q; 31 int solve(int l,int k) 32 { 33 if(l<k)return 0; 34 else return 1+(l-k)/(k+1); 35 } 36 int main() 37 { 38 #ifdef Local 39 freopen("data.txt","r",stdin); 40 #endif 41 int n,k,a,m,p,sum=0,flag=0,ans; 42 cin>>n>>k>>a; 43 cin>>m; 44 for(int i=1;i<=m;i++)scanf("%d",&x[i]); 45 Q.insert(0);Q.insert(n+1); 46 sum=solve(n,a); 47 for(int i=1;i<=m;i++) 48 { 49 set<int>::iterator it1,it2,it3,it; 50 Q.insert(x[i]); 51 it1=Q.find(x[i]); 52 it1--; 53 it2=Q.find(x[i]); 54 it2++; 55 sum=sum-solve(*(it2)-*(it1)-1,a)+solve(x[i]-*(it1)-1,a)+solve(*(it2)-x[i]-1,a); 56 if(sum<k){ 57 flag=i; 58 break; 59 } 60 61 } 62 if(flag)cout<<flag<<endl; 63 else cout<<-1<<endl; 64 #ifdef Local 65 cerr << "time: " << (LL) clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl; 66 #endif 67 }