Problem 12 : Highly divisible triangular number

Problem 12

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Highly divisible triangular number

The sequence of triangle numbers is generated by adding the natural numbers. So the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The first ten terms would be:

1, 3, 6, 10, 15, 21, 28, 36, 45, 55, …

Let us list the factors of the first seven triangle numbers:

 1: 1
 3: 1,3
 6: 1,2,3,6
10: 1,2,5,10
15: 1,3,5,15
21: 1,3,7,21
28: 1,2,4,7,14,28

We can see that 28 is the first triangle number to have over five divisors.

What is the value of the first triangle number to have over five hundred divisors?

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高度可约的三角形数

三角形数数列是通过逐个加上自然数来生成的。例如，第7个三角形数是 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28。三角形数数列的前十项分别是：

1, 3, 6, 10, 15, 21, 28, 36, 45, 55, …

让我们列举出前七个三角形数的所有约数：

 1: 1
 3: 1,3
 6: 1,2,3,6
10: 1,2,5,10
15: 1,3,5,15
21: 1,3,7,21
28: 1,2,4,7,14,28

我们可以看出，28是第一个拥有超过5个约数的三角形数。

第一个拥有超过500个约数的三角形数是多少？

题目解答

    朴素解法：暴力枚举

   优化算法：利用线性筛法计算计算每个数字的因子个数

    

题目代码

#include <stdio.h>
#include <inttypes.h>
#include <stdlib.h>
#define N 2000000
int32_t prime[N + 5] = {0};
int32_t factor[N + 5] = {0};
void init(){
    for (int i = 2; i < N; i++){
        if (!prime[i]){
            prime[++prime[0]] = i;
            factor[i] = 2;
        }
        for (int j = 1; j <= prime[0]; j++){
            if (i * prime[j] > N) break;
            prime[i * prime[j]] = 1;
            if(i % prime[j] == 0){
                int cnt = 0;
                int temp = i;
                while(temp % prime[j] == 0){
                    temp /= prime[j];
                    ++cnt;
                }
                factor[i * prime[j]] = factor[i] / (cnt + 1) * (cnt + 2);
                break;
            }else{
                factor[i * prime[j]] = factor[i] * factor[prime[j]];
            }
        }
    }
}
int main() {
    init();
    for (int i=1;;i++){
        int temp;
        if(i & 1){
            temp = factor[(i+1)/2] * factor[i];
        }else{
            temp = factor[i+1] * factor[i/2];
        }
        if(temp > 500){
            printf("%d\n",i*(i+1)/2);
            break;
        }
    }
    return 0;
}