标签：target ram val ase ems end you int lld
There are nn boxers, the weight of the ii-th boxer is aiai. Each of them can change the weight by no more than 11 before the competition (the weight cannot become equal to zero, that is, it must remain positive). Weight is always an integer number.
It is necessary to choose the largest boxing team in terms of the number of people, that all the boxers‘ weights in the team are different (i.e. unique).
Write a program that for given current values ?aiai will find the maximum possible number of boxers in a team.
It is possible that after some change the weight of some boxer is 150001150001 (but no more).
The first line contains an integer nn (1≤n≤1500001≤n≤150000) — the number of boxers. The next line contains nn integers a1,a2,…,ana1,a2,…,an, where aiai (1≤ai≤1500001≤ai≤150000) is the weight of the ii-th boxer.
Print a single integer — the maximum possible number of people in a team.
4
3 2 4 1
4
6
1 1 1 4 4 4
5
In the first example, boxers should not change their weights — you can just make a team out of all of them.
In the second example, one boxer with a weight of 11 can be increased by one (get the weight of 22), one boxer with a weight of 44 can be reduced by one, and the other can be increased by one (resulting the boxers with a weight of 33 and 55, respectively). Thus, you can get a team consisting of boxers with weights of 5,4,3,2,15,4,3,2,1.
#include <bits/stdc++.h> #include <iostream> #include <algorithm> #include <cstdio> #include <string> #include <cstring> #include <cstdlib> #include <map> #include <vector> #include <set> #include <queue> #include <stack> #include <cmath> typedef long long ll; using namespace std; const int INT=1e6+5; #define lson rt<<1, l, m #define rson rt<<1|1, m+1, r #define read(x) scanf("%d",&x) #define lread(x) scanf("%lld",&x); #define pt(x) printf("%d\n",(x)) #define cn cin>> #define ct cout<< #define en <<endl #define rep(j,k) for (int i = (int)(j); i <= (int)(k); i++) #define input(k) for (int i = 1; i <= (int)(k); i++) {cin>>a[i] ; } #define mem(s,t) memset(s,t,sizeof(s)) #define re return 0; #define TLE std::ios::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL); int a[151000]; bool vis[151000]; int main() { TLE; int n; scanf("%d",&n); for(int i = 1; i <= n; ++i) scanf("%d",&a[i]); sort(a+1,a+1+n); int sum = 0; for(int i = 1; i <= n; ++i){ if(!vis[a[i]-1] && a[i] != 1){ vis[a[i]-1] = true; sum++;continue; } if(!vis[a[i]]){ vis[a[i]] = true; sum++;continue; } if(!vis[a[i]+1]){ vis[a[i]+1] = true; sum++;continue; } } cout<<sum<<endl; return 0; }
E. Boxers （ Codeforces Round #579 ）
标签：target ram val ase ems end you int lld
原文地址：https://www.cnblogs.com/Shallow-dream/p/11416516.html
