Title
Growth Of Population
Description
In a small town the population is p0 = 1000 at the beginning of a year. The population regularly increases by 2 percent per year and moreover 50 new inhabitants per year come to live in the town. How many years does the town need to see its population greater or equal to p = 1200 inhabitants?
At the end of the first year there will be:
1000 + 1000 * 0.02 + 50 => 1070 inhabitants
At the end of the 2nd year there will be:
1070 + 1070 * 0.02 + 50 => 1141 inhabitants (number of inhabitants is an integer)
At the end of the 3rd year there will be:
1141 + 1141 * 0.02 + 50 => 1213
It will need 3 entire years.
More generally given parameters:
p0, percent, aug (inhabitants coming or leaving each year), p (population to surpass)
the function nb_year should return n number of entire years needed to get a population greater or equal to p.
aug is an integer, percent a positive or null number, p0 and p are positive integers (> 0)
Examples:
nb_year(1500, 5, 100, 5000) -> 15
nb_year(1500000, 2.5, 10000, 2000000) -> 10
How Can I Solved
문제 요구사항 정의
p 이상의 돈을 모으는데 몇년이나 걸리는지를 구하는 문제이다.
문제 접근
매년 원금(p0)에서 일정 퍼센트(p0*percent/100) + 보너스(aug)만큼 오른다.
재귀 함수를 사용해서 재귀가 한번 호출 될때마다 count++을 해준다.
p0=(p0+p0*percent/100 +aug)를 함으로써
계속해서 p0를 초기화 시켜준다.
Solution
function nbYear(p0, percent, aug, p,count=0) {
return p0>=p ? count :(count+=1, nbYear(p0=(p0+ p0* percent/100 +aug),percent,aug,p,count))
}
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