求r中给定数的所有2个幂值组合值的和

Question

R数据帧1:

tbody> <<tr>1245

Index	Powervalue
0	1
	2
	4
3	8
	16
	32

这是一个使用intToBits转换为二进制库和tidyverse库的解决方案，用于后续的数据操作。

x <- c(20L, 50L)
map_dfr(x, ~{
bin <- intToBits(.x)
n <- 2^(seq_len(32) - 1)
possible <- paste(n[as.logical(bin)], collapse = ", ")
data.frame(combinevalue = .x,
possiblecodes = possible,
y = rev(as.integer(bin)),
n = rev(paste0("code_", n)))
}) |>
pivot_wider(names_from = n, values_from = y)

##> # A tibble: 2 × 34
##>   combinevalue possiblecodes code_2147483648 code_1073741824 code_536870912
##>          <int> <chr>                   <int>           <int>          <int>
##> 1           20 4, 16                       0               0              0
##> 2           50 2, 16, 32                   0               0              0
##> # … with 29 more variables: code_268435456 <int>, code_134217728 <int>,
##> #   code_67108864 <int>, code_33554432 <int>, code_16777216 <int>,
##> #   code_8388608 <int>, code_4194304 <int>, code_2097152 <int>,
##> #   code_1048576 <int>, code_524288 <int>, code_262144 <int>,
##> #   code_131072 <int>, code_65536 <int>, code_32768 <int>, code_16384 <int>,
##> #   code_8192 <int>, code_4096 <int>, code_2048 <int>, code_1024 <int>,
##> #   code_512 <int>, code_256 <int>, code_128 <int>, code_64 <int>, …

尝试下面的递归

f <- function(n) {
if (n == 0) {
return(NULL)
}
v <- 2^floor(log2(n))
c(Recall(n - v), v)
}

得到

> f(50)
[1]  2 16 32
> f(20)
[1]  4 16