D1. Coffee and Coursework (Easy version) and D2. Coffee and Coursework (Hard Version)

http://codeforces.com/contest/1118/problem/D1

http://codeforces.com/contest/1118/problem/D2

题意：有n杯咖啡，m张作业，没杯咖啡有能量点a[i]，一点能量能做一张作业，同一天喝多杯能量会减少：第一杯a[i],第二杯max(0,a[i]-1),第三杯max(0,a[i]-2).....；问完成m张作业最少天数，不能完成-1；

思路：贪心，从最大的开始取；二分，设k天完成，第一次k天就是a[i],第二次k天a[i]-1，第三个a[i]-2(相当于把后面的一个个加到第一个k天)....

#include<algorithm>
#include<set>
#include<vector>
#include<queue>
#include<cmath>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<set>
#include<vector>
#include<queue>
#include<cmath>
#include<cstring>
#include<sstream>
#include<cstdio>
#include<ctime>
#include<map>
#include<stack>
#include<string>
using namespace std;#define sfi(i) scanf("%d",&i)
#define pri(i) printf("%d\n",i)
#define sff(i) scanf("%lf",&i)
#define ll long long
#define mem(x,y) memset(x,y,sizeof(x))
#define INF 0x3f3f3f3f
#define eps 1e-6
#define PI acos(-1)
#define lowbit(x) ((x)&(-x))
#define zero(x) (((x)>0?(x):-(x))<eps)
#define fl() printf("flag\n")
ll gcd(ll a,ll b){while(b^=a^=b^=a%=b);return a;}
const int maxn=2e5+9;
const int mod=1e9+7;template <class T>
inline void sc(T &ret)
{char c;ret = 0;while ((c = getchar()) < '0' || c > '9');while (c >= '0' && c <= '9'){ret = ret * 10 + (c - '0'), c = getchar();}
}
ll a[maxn];
ll n,m;
bool ch(int mid)
{ll sum=0;ll num=0;ll sb=0;for(int i=0;i<n;i++){sum=sum+max((ll)0,a[i]-sb);num++;if(num==mid) sb++,num=0;}return sum>=m;
}int main()
{cin>>n>>m;for(int i=0;i<n;i++){cin>>a[i];}sort(a,a+n);reverse(a,a+n);int l=1;int r=n;int ans=-1;while(l<=r){int mid=(l+r)>>1;if(ch(mid)){r=mid-1;ans=mid;}else l=mid+1;}cout<<ans<<endl;return 0;
}

python:

maxn=200000+9def ch(mid):sum=0num=0sb=0for i in range(n):sum+=max(0,int(a[i]-sb))num+=1if num==mid:sb+=1num=0return sum>=mop=input()
op=op.split()
n=int(op[0])
m=int(op[1])a=[0]*nop=input()
op=op.split()for i in range(len(op)):a[i]=int(op[i])a.sort()
a.reverse()l=1
r=n
ans=-1
while l<=r:mid=(l+r)>>1if ch(mid):ans=midr=mid-1else :l=mid+1
print(ans)

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