10.05T2 二项式展开+二分答案+差分数组

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cmath> 
 4 #include<cstring>
 5 #define N 500005
 6 using namespace std;
 7 const long long inf=1e18;
 8 long long read(){
 9     long long s=0,f=1;
10     char t=getchar();
11     while(t<‘0‘||t>‘9‘){
12         if(t==‘-‘)f=-1;
13         t=getchar();
14     }
15     while(t>=‘0‘&&t<=‘9‘){
16         s=(s<<1)+(s<<3)+t-‘0‘;
17         t=getchar();
18     }
19     return s*f;
20 }
21 long long ksm(long long a,long long b){
22     long long ans=1;
23     while(b){
24         if(b&1)ans=ans*a;
25         b>>=1;a=a*a;
26     }
27     if(ans<0)return inf;
28     return ans;
29 }
30 long long c[30][30],n,m,k;
31 void pre(){
32     c[0][0]=1;
33     for(long long i=1;i<=3;i++){
34         c[i][i]=c[i][0]=1;
35         for(long long j=1;j<i;j++)c[i][j]=c[i-1][j]+c[i-1][j-1];
36     }
37 }
38 long long add[N],val[N],Pow[N],S[N],T[N][4],siz[N];
39 bool check(long long p){
40     long long used=0,Siz=0,sum,dis=ceil(pow(p,1.0/m));
41     if(m>3){
42         memset(add,0,sizeof add);
43         for(long long i=n,x;i>=1;i--){
44             if(add[i]>val[i])continue;
45             x=(val[i]-add[i])/p+1;
46             used+=x;if(used>k)return false;
47             for(long long j=i-1;j>=1&&Pow[i-j]<p;j--){
48                 add[j]+=x*(p-Pow[i-j]);
49             }
50         }
51     }
52     else{
53         memset(S,0,sizeof S);
54         memset(T,0,sizeof T);
55         memset(siz,0,sizeof siz);
56         for(long long i=n,x,now;i>=1;i--){
57             Siz+=siz[i];sum=Siz*p;now=1;
58             for(long long j=0;j<=m;j++,now*=i){
59                 S[j]+=T[i][j];
60                 sum-=now*S[j];
61             }
62             if(sum>val[i])continue;
63             x=(val[i]-sum)/p+1;used+=x;
64             if(used>k)return false;
65             long long lim=i-dis,fla=1;now=ksm(i,m);lim=max(lim,0ll);
66             for(long long j=0;j<=m;j++,fla*=-1,now/=i){
67                 T[lim][j]-=fla*x*c[m][j]*now;
68                 S[j]+=x*fla*c[m][j]*now;
69             }
70             Siz+=x;siz[lim]-=x;
71         }
72     }
73     return true;
74 }
75 long long ans;
76 int main(){
77     n=read(),m=read(),k=read();
78     pre();
79     for(long long i=1;i<=n;i++)val[i]=read();
80     for(long long i=1;i<=n;i++)Pow[i]=ksm(i,m);
81     long long l=1,r=1000000;
82     while(l<=r){
83         long long mid=l+r>>1;
84         if(check(mid))r=mid-1,ans=mid;
85         else l=mid+1;
86     }
87     cout<<ans;
88     return 0;
89 }