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\begin{aligned} s^2&=\frac{1}{k}\sum_{i=1}^k(\bar{v}-v_i)^2s\\ &=\frac 1k\Big( \sum_{i=1}^k (\bar v^2+v_i^2-2\bar vv_i)\Big)\\ &=\frac 1k\Big( \sum_{i=1}^k \bar v^2+\sum_{i=1}^kv_i^2-\sum_{i=1}^k2\bar vv_i)\Big)\\ &=\frac 1k\Big( k\bar v^2+\sum_{i=1}^kv_i^2-2\bar v\sum_{i=1}^kv_i)\Big)\\ &=\bar v^2+\frac 1k\sum_{i=1}^kv_i^2-\frac 2k\bar v\sum_{i=1}^kv_i\\ &=\frac 1{k^2}\Big(\sum_{i=1}^k v_i\Big)^2+\frac 1k\sum_{i=1}^kv_i^2-\frac 2k\bar v\sum_{i=1}^kv_i\\ &=\frac 1{k^2}\Big(\sum_{i=1}^k v_i\Big)^2+\frac 1k\sum_{i=1}^kv_i^2-\frac 2{k^2}\Big(\sum_{i=1}^kv_i\Big)^2\\ &=\frac 1k\sum_{i=1}^kv_i^2-\frac 1{k^2}\Big(\sum_{i=1}^kv_i\Big)^2\\ \end{aligned}
s 2 = k 1 i = 1 ∑ k ( v ˉ − v i ) 2 s = k 1 ( i = 1 ∑ k ( v ˉ 2 + v i 2 − 2 v ˉ v i ) ) = k 1 ( i = 1 ∑ k v ˉ 2 + i = 1 ∑ k v i 2 − i = 1 ∑ k 2 v ˉ v i ) ) = k 1 ( k v ˉ 2 + i = 1 ∑ k v i 2 − 2 v ˉ i = 1 ∑ k v i ) ) = v ˉ 2 + k 1 i = 1 ∑ k v i 2 − k 2 v ˉ i = 1 ∑ k v i = k 2 1 ( i = 1 ∑ k v i ) 2 + k 1 i = 1 ∑ k v i 2 − k 2 v ˉ i = 1 ∑ k v i = k 2 1 ( i = 1 ∑ k v i ) 2 + k 1 i = 1 ∑ k v i 2 − k 2 2 ( i = 1 ∑ k v i ) 2 = k 1 i = 1 ∑ k v i 2 − k 2 1 ( i = 1 ∑ k v i ) 2
则
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s^2=\frac 1m\sum_{i=1}^mv_i^2-\frac 1{m^2}\Big(\sum_{i=1}^mv_i\Big)^2
s 2 = m 1 i = 1 ∑ m v i 2 − m 2 1 ( i = 1 ∑ m v i ) 2
乘上
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s^2m^2=m\sum_{i=1}^mv_i^2-\Big(\sum_{i=1}^mv_i\Big)^2
s 2 m 2 = m i = 1 ∑ m v i 2 − ( i = 1 ∑ m v i ) 2
由于
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\Big(\sum\limits_{i=1}^mv_i\Big)^2
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sum[i]=\sum\limits_{i=1}^i a_i
s u m [ i ] = i = 1 ∑ i a i
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f[i][j]=\min\limits_{k=1}^{i-1}f[k][j-1]+(sum[i]-sum[k])^2
f [ i ] [ j ] = k = 1 min i − 1 f [ k ] [ j − 1 ] + ( s u m [ i ] − s u m [ k ] ) 2
Θ
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\Theta(n^3)
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f[k_a][j-1]+(sum[i]-sum[k_a])^2<f[k_b][j-1]+(sum[i]-sum[k_b])^2
f [ k a ] [ j − 1 ] + ( s u m [ i ] − s u m [ k a ] ) 2 < f [ k b ] [ j − 1 ] + ( s u m [ i ] − s u m [ k b ] ) 2
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f[k_a][j-1]+2sum[i]*sum[k_a]+sum[k_a]^2<f[k_b][j-1]+2sum[i]*sum[k_b]+sum[k_b]^2
f [ k a ] [ j − 1 ] + 2 s u m [ i ] ∗ s u m [ k a ] + s u m [ k a ] 2 < f [ k b ] [ j − 1 ] + 2 s u m [ i ] ∗ s u m [ k b ] + s u m [ k b ] 2
化简：
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(f[k_a][j-1]+sum[k_a]^2)-(f[k_b][j-1]+sum[k_b]^2)<2sum[i]\times(sum[k_a]-sum[k_b])
( f [ k a ] [ j − 1 ] + s u m [ k a ] 2 ) − ( f [ k b ] [ j − 1 ] + s u m [ k b ] 2 ) < 2 s u m [ i ] × ( s u m [ k a ] − s u m [ k b ] )
整理：整理：
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( s u m [ k a ] − s u m [ k b ] ) ( f [ k a ] [ j − 1 ] + s u m [ k a ] 2 ) − ( f [ k b ] [ j − 1 ] + s u m [ k b ] 2 ) < 2 s u m [ i ]
单调队列维护递增斜率即可，转移时间复杂度
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