time limit per test : 1 secondios
memory limit per test : 256 megabytesc++
input : standard inputspa
output : standard outputcode
Alice and Bob love playing one-dimensional battle ships. They play on the field in the form of a line consisting of \(n\) square cells (that is, on a \(1 ×n\) table).orm
At the beginning of the game Alice puts \(k\) ships on the field without telling their positions to Bob. Each ship looks as a \(1 × a\) rectangle (that is, it occupies a sequence of \(a\) consecutive squares of the field). The ships cannot intersect and even touch each other.排序
After that Bob makes a sequence of "shots". He names cells of the field and Alice either says that the cell is empty ("miss"), or that the cell belongs to some ship ("hit").three
But here's the problem! Alice like to cheat. May be that is why she responds to each Bob's move with a "miss".ip
Help Bob catch Alice cheating — find Bob's first move, such that after it you can be sure that Alice cheated.ci
The first line of the input contains three integers: \(n, k\) and \(a (1 ≤ n, k, a ≤ 2·10^5)\) — the size of the field, the number of the ships and the size of each ship. It is guaranteed that the \(n, k\) and \(a\) are such that you can put \(k\) ships of size \(a\) on the field, so that no two ships intersect or touch each other.input
The second line contains integer \(m (1 ≤ m ≤ n)\) — the number of Bob's moves.
The third line contains \(m\) distinct integers \(x_1, x_2, ..., x_m\), where \(x_i\) is the number of the cell where Bob made the \(i\)-th shot. The cells are numbered from left to right from \(1\) to \(n\).
Print a single integer — the number of such Bob's first move, after which you can be sure that Alice lied. Bob's moves are numbered from \(1\) to \(m\) in the order the were made. If the sought move doesn't exist, then print "\(-1\)".
input
11 3 3 5 4 8 6 1 11
output
3
input
5 1 3 2 1 5
output
-1
input
5 1 3 1 3
output
1
一个长为\(1\times n\)的区域内有\(k\)个战舰,每一个战舰的长度均为\(a\),Bob射击\(m\)个不一样的位置,对于每次射击,Alice都会告诉Bob是否击中。
可是由于Alice会撒谎,因此当Bob击中的时候,Alice也会说没有击中
问Bob最先能够在第几回射击的时候发现Alice在说谎
由于随着射击次数的增多,Bob发现Alice说谎的可能性越大,因此能够利用二分来解决
将\([1,mid]\)区间的位置进行排序,而后计算这些位置若是所有没有击中,能够放下多少战舰,若是超过\(k\)个,那么当前的\(mid\)是可行的,查询前半部分,不然,查询后半部分
#include <bits/stdc++.h> #define ll long long #define ull unsigned long long #define ms(a,b) memset(a,b,sizeof(a)) const int inf=0x3f3f3f3f; const ll INF=0x3f3f3f3f3f3f3f3f; const int maxn=1e6+10; const int mod=1e9+7; const int maxm=1e3+10; using namespace std; int x[maxn]; int b[maxn]; int n,k,a; bool check(int mid) { for(int i=1;i<=mid;i++) b[i]=x[i]; sort(b+1,b+1+mid); int cnt=0; for(int i=1;i<=mid;i++) cnt+=(b[i]-b[i-1])/(a+1); // b[i]和b[i-1]之间能够放多少战舰 // b[mid]->最后一个位置能够放多少战舰 cnt+=(n-b[mid]+1)/(a+1); if(cnt>=k) return true; return false; } int main(int argc, char const *argv[]) { #ifndef ONLINE_JUDGE freopen("/home/wzy/in", "r", stdin); freopen("/home/wzy/out", "w", stdout); srand((unsigned int)time(NULL)); #endif ios::sync_with_stdio(false); cin.tie(0); cin>>n>>k>>a; int m; cin>>m; for(int i=1;i<=m;i++) { cin>>x[i]; } int l=1,r=m,ans=inf; while(l<=r) { int mid=(l+r)/2; if(check(mid)) l=mid+1; else r=mid-1,ans=min(ans,mid); } if(ans==inf) cout<<-1<<endl; else cout<<ans<<endl; #ifndef ONLINE_JUDGE cerr<<"Time elapsed: "<<1.0*clock()/CLOCKS_PER_SEC<<" s."<<endl; #endif return 0; }