2019-2020 ACM-ICPC Brazil Subregional Programming Contest Problem M Maratona Brasileira de Popcorn (二分)

• 题意:有\(n\)袋爆米花,某个队伍有\(c\)个队员,每个队员每秒做多可以吃\(t\)粒爆米花,但一袋爆米花只能由一个队员吃完,并且一个队员只能吃连续的一袋或几袋,不能隔着吃某一袋,求将所有爆米花吃完的最少时间.

• 题解:这道题当时想了半天,发现怎么也求不出答案,后来想到了二分答案的办法,将答案代入并模拟题意判断看是否满足条件,二分求得最小值即可.

• 代码:

  #include <iostream>
  #include <cstdio>
  #include <cstring>
  #include <cmath>
  #include <algorithm>
  #include <stack>
  #include <queue>
  #include <vector>
  #include <map>
  #include <set>
  #include <bitset>
  #include <unordered_set>
  #include <unordered_map>
  #define ll long long
  #define fi first
  #define se second
  #define pb push_back
  #define me memset
  const int N = 1e6 + 10;
  const int mod = 1e9 + 7;
  const int INF = 0x3f3f3f3f;
  using namespace std;
  typedef pair<int,int> PII;
  typedef pair<ll,ll> PLL;
   
  int n,c,t;
  ll a[N];
   
  bool check(ll x){
  	ll num=1;
  	ll sum=0;
  	ll now=x*t;
  	for(int i=1;i<=n;++i){
  		if(a[i]>now){
  			return false;
  		}
  		if(sum+a[i]>now){
  			num++;
  			sum=a[i];
  		}
  		else{
  			sum+=a[i];
  		}
  	}
  	if(num<=c) return true;
  	else return false;
  }
   
  int main() {
      ios::sync_with_stdio(false);cin.tie(0);
  	cin>>n>>c>>t;
  	for(int i=1;i<=n;++i){
  		cin>>a[i];
  	}
  	ll l=1,r=INF;
  	while(l<r){                 //模板:向左二分
  		ll mid=l+r>>1;
  		if(check(mid)) r=mid;
  		else l=mid+1;
  	}
  	cout<<r<<endl;
  	
      return 0;
  }