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法律研究质数
质数是指质因子为 2、3 或 5 的数。从 1 到 15，有 11 个质数：1、2、3、4、5、6、8、9、10、12、15。7、11、13 不是质数，因为它们是质数。14 不是质数，因为它的质因子中包含 7。
在这个程序中，我们将尝试找出第 n 个质数。
输入和输出
Input: Take the term number. Say it is 10 Output: The 10th ugly number is 12
算法
getUglyNumbers(n)
输入：项数。
输出：找到第 n 个质数。
Begin define array named uglyNum of size n i2 := 0, i3 := 0, i5 := 0 next2mul := 2, next3mul := 3, next5Mul := 5 next := 1 ugluNum[0] := 1 for i := 1 to n, do next := minimum of next2Mul, next3Mul and next5Mul uglyNum[i] := next if next = next2Mul, then i2 := i2 + 1 next2mul := uglyNum[i2] * 2 if next = next3Mul, then i3 := i3 + 1 next3mul := uglyNum[i3] * 3 if next = next5Mul, then i5 := i5 + 1 next5mul := uglyNum[i5] * 5 done return next End
示例
# include<iostream>
using namespace std;
int min(int x, int y, int z) { //find smallest among three numbers
if(x < y) {
if(x < z)
return x;
else
return z;
}else {
if(y < z)
return y;
else
return z;
}
}
int getUglyNum(int n) {
int uglyNum[n]; // To store ugly numbers
int i2 = 0, i3 = 0, i5 = 0;
//find next multiple as 1*2, 1*3, 1*5
int next2mul = 2;
int next3mul = 3;
int next5mul = 5;
int next = 1; //initially the ugly number is 1
uglyNum[0] = 1;
for (int i=1; i<n; i++) {
next = min(next2mul, next3mul, next5mul); //find next ugly number
uglyNum[i] = next;
if (next == next2mul) {
i2++; //increase iterator of ugly numbers whose factor is 2
next2mul = uglyNum[i2]*2;
}
if (next == next3mul) {
i3++; //increase iterator of ugly numbers whose factor is 3
next3mul = uglyNum[i3]*3;
}
if (next == next5mul) {
i5++; //increase iterator of ugly numbers whose factor is 5
next5mul = uglyNum[i5]*5;
}
}
return next; //the nth ugly number
}
int main() {
int n;
cout << "Enter term: "; cin >> n;
cout << n << "th Ugly number is: " << getUglyNum(n) << endl;
}输出
Enter term: 10 10th Ugly number is: 12
更新日期： 17-6-2020
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