HDU1299 Diophantus of Alexandria 素因子分解

一个需要进行转化的题目。

x = (n*(n+k))/k = n*n/k + n,最后求小于等于N且是N^2的因子的数的个数。

代码如下:

#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <map>
using namespace std;

int N;
map<int,int>mp;
map<int,int>::iterator it;

int fun()
{
    int ret = 1;
    for (it = mp.begin(); it != mp.end(); ++it) {
        ret *= (1 + 2 * it->second);
    }
    return ret;
}

int main()
{
    int ca = 0, T, lim, sum;
    scanf("%d", &T);
    while (T--) {
        sum = 0;
        mp.clear();
        scanf("%d", &N);
        lim = (int)sqrt(N*1.);
        for (int i = 2; i <= lim; ++i) {
            while (N % i == 0) {
                ++mp[i];
                N /= i;
            }
        }
        printf("Scenario #%d:\n", ++ca);
        sum = fun();
        if (N != 1) {
            sum *= 3;
        }
        printf("%d\n\n", sum + 1 >> 1);
    }
    return 0;
}
原文地址:https://www.cnblogs.com/Lyush/p/2608030.html