计算机数学和算法习题和解答(1~3)

1. If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.

Find the sum of all the multiples of 3 or 5 below 1000.

找到小于1000的 而且 是3或者5的倍数的 自然数，求它们的和。

x <- 1

sum <- 0

while (x < 1000) {

if (x%%3==0 | x%%5==0) {

sum <- sum + x

}

x <- x + 1

}

print(sum)

2. Each new term in the Fibonacci sequence(http://baike.baidu.com/view/1074762.htm) is generated by adding the previous
two terms. By starting with 1 and 2, the first 10 terms will be:

1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...
By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.

有一个小于400万的斐波那契数列，求其中的偶数和。

i <- 2

x <- 1:2

while (x[i]<4000000) {

x[i+1] <- x[i-1]+x[i]

i <- i+1

}

sum(x[x%%2==0])

3. The prime factors of 13195 are 5, 7, 13 and 29.

What is the largest prime factor of the number 600851475143 ?

求600851475143的最大质数因子。
findprime <- function(x) {

if (x %in% c(2,3,5,7)) return(TRUE)

if (x%%2 == 0 | x==1) return(FALSE)

xsqrt <- round(sqrt(x))

xseq <- seq(from=3,to=xsqrt,by=2)

if (all(x %% xseq !=0)) return(TRUE)

else return(FALSE)

}

# x = 1:111

# x[sapply(x,findprime)]

maxfactor <- 0

n <- 110

for (i in seq(from=3, to=round(sqrt(n))+1, by=2)) {

if (findprime(i) & (n %% i == 0)) {

#i是质数因子

n <- n / i

maxfactor <- i

if (i >= n) break

}

}

print(maxfactor)

参考文章求最大质因子的N种境界