Chi Shen, Oct 12 2019
题目描述
质数的定义为：在大于1的自然数中，除了1和它本身，不能整除其他自然数的数，如2，3，23等。
给定一个取值范围[low, high]，找到该范围内的所有质数，输出这些质数的十位数之和与个位数之和的较小值，如果该质数小于10，则其十位数为0，注意low在取值范围内，high不在。
输入描述
第一行输入：low，high，用例保证均不小于0，且low < high，不用考虑输入合法性，如果没有质数，则输出0。
输出描述
输出找到的质数和十位数之和与个位数之和的较小值。
解题思路
# Rversion: 3.5.3
lower <- as.numeric(readline('Step 1：Please input a smaller integer value larger than 1:'))
upper <- as.numeric(readline('Step 2：Please input a larger integer value larger than 1:'))
if (lower >= upper) {
print('Please make sure the second value is larger than first value！')
} else {
sequence <- seq(lower, upper - 1, 1)
#---------------判断质数------------------#
x <- vector(mode = 'logical', length = upper - lower)
for (i in 1:length(x)) {
if (all(sequence[i] %% (2:sqrt(sequence[i])) != 0)) {
x[i] <- TRUE
} else {
x[i] <- FALSE
}
}
prime_num <- sequence[x]
if (length(prime_num) == 0) {
sprintf('There is no prime number between %i and %i', lower, upper)
} else {
#---------------取个位数和十位数------------------#
ten_dig <- vector(mode = 'numeric', length = length(prime_num))
sin_dig <- vector(mode = 'numeric', length = length(prime_num))
for (j in 1:length(prime_num)) {
ten_dig[j] <- prime_num[j] %/% 10 %% 10 #取除数再取余数
sin_dig[j] <- prime_num[j] %% 10 #取余数
}
#---------------分别求个位数和十位数之和------------------#
ten_sum <- sum(ten_dig)
sin_sum <- sum(sin_dig)
sprintf('The small value of summarize of ten digits (%i) and summarize of single digits (%i) in prime number in [%i，%i）is: %i',
ten_sum, sin_sum, lower, upper, min(c(ten_sum, sin_sum)))
}
}