题目链接
题目大意：给你一个长度为n的数组，你可以选择一段区间将这段区间的数全都变成这段区间的平均值，问你最后这个数组字典序最小是怎么样的
解题思路：1.首先我们知道最后这个序列一定会变成一个单调上升的子序列 2.然后我们发现操作之后序列会变成一块一块的，每一块的数字都是相同的，那么我们可以按照分块的思想，如果这一块平均值比前面的小就合并到前面去
#include <iostream>
#include <cstdio>
#include <stack>
#include <sstream>
#include <vector>
#include <map>
#include <cstring>
#include <deque>
#include <cmath>
#include <iomanip>
#include <queue>
#include <algorithm>
#include <set>
#define mid ((l + r) >> 1)
#define Lson rt << 1, l , mid
#define Rson rt << 1|1, mid + 1, r
#define ms(a,al) memset(a,al,sizeof(a))
#define log2(a) log(a)/log(2)
#define _for(i,a,b) for( int i = (a); i < (b); ++i)
#define _rep(i,a,b) for( int i = (a); i <= (b); ++i)
#define for_(i,a,b) for( int i = (a); i >= (b); -- i)
#define rep_(i,a,b) for( int i = (a); i > (b); -- i)
#define lowbit(x) ((-x) & x)
#define IOS std::ios::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define INF 0x3f3f3f3f
#define hash Hash
#define next Next
#define count Count
#define pb push_back
#define Setw(a) fixed<<setprecision(a)
#define f first
#define s second
using namespace std
;
const int N
= 1e6+10, mod
= 1e9 + 7;
const double eps
= 1e-10;
typedef long long ll
;
typedef unsigned long long ull
;
typedef pair
<int,int> PII
;
template<typename T
> void read(T
&x
)
{
x
= 0;char ch
= getchar();ll f
= 1;
while(!isdigit(ch
)){if(ch
== '-')f
*=-1;ch
=getchar();}
while(isdigit(ch
)){x
= x
*10+ch
-48;ch
=getchar();}x
*=f
;
}
template<typename T
, typename... Args
> void read(T
&first
, Args
& ... args
)
{
read(first
);
read(args
...);
}
double a
[N
], ans
[N
];
int block
[N
], l
;
int main()
{
int T
;
read(T
);
_for(i
,1,T
+1) read(a
[i
]);
_for(i
,1,T
+1)
{
ans
[++ l
] = a
[i
];
block
[l
] = 1;
while(l
> 1 && ans
[l
] < ans
[l
- 1])
{
ans
[l
- 1] = (ans
[l
] * block
[l
] + ans
[l
- 1] * block
[l
- 1]) / (block
[l
] + block
[l
- 1]);
block
[l
- 1] = block
[l
- 1] + block
[l
];
l
--;
}
}
_for(i
,1,l
+1)
_for(j
,1,block
[i
]+1)
cout
<<Setw(10) << ans
[i
] << "\n";
return 0;
}
