CSP-202312-2-因子化简
解题思路
本题要求实现的是素数分解，并检查每个素因子的指数是否大于等于
k，满足条件则将其加入最终乘积中，最后输出这个乘积。如果没有任何素因子的指数大于等于k，则按照题目要求输出1。
-
输入测试用例数
q，对于每个测试用例，首先读入num和k。 -
初始化一个
numberList，用于存储满足条件（素因子的指数大于等于k）的素数及其指数。 -
遍历一个已知的素数列表
primeList，对于每个素数，检查它是否为num的因子。如果是，则计算该素数的指数ti，即该素数能连续整除num的次数，直到num不能被该素数整除。 -
如果某个素数的指数
ti大于等于k，则将该素数及其指数作为一个结构体MyNumber加入到numberList中。 -
最后，检查
numberList。如果numberList为空，即没有任何素数的指数大于等于k，则输出1。否则，遍历numberList，计算所有元素对应的pi^ei（即素数及其指数的乘积）的总乘积，并输出这个总乘积。
注意：本题涉及的数字都比较大，建议使用long long定义整形数据。
完整代码
#include<iostream>
#include<vector>
#include <cmath>
using namespace std;
// 5000以内的素数
vector<int>primeList = {
2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197,199,211,223,227,229,233,239,241,251,257,263,269,271,277,281,283,293,307,311,313,317,331,337,347,349,353,359,367,373,379,383,389,397,401,409,419,421,431,433,439,443,449,457,461,463,467,479,487,491,499,503,509,521,523,541,547,557,563,569,571,577,587,593,599,601,607,613,617,619,631,641,643,647,653,659,661,673,677,683,691,701,709,719,727,733,739,743,751,757,761,769,773,787,797,809,811,821,823,827,829,839,853,857,859,863,877,881,883,887,907,911,919,929,937,941,947,953,967,971,977,983,991,997,1009,1013,1019,1021,1031,1033,1039,1049,1051,1061,1063,1069,1087,1091,1093,1097,1103,1109,1117,1123,1129,1151,1153,1163,1171,1181,1187,1193,1201,1213,1217,1223,1229,1231,1237,1249,1259,1277,1279,1283,1289,1291,1297,1301,1303,1307,1319,1321,1327,1361,1367,1373,1381,1399,1409,1423,1427,1429,1433,1439,1447,1451,1453,1459,1471,1481,1483,1487,1489,1493,1499,1511,1523,1531,1543,1549,1553,1559,1567,1571,1579,1583,1597,1601,1607,1609,1613,1619,1621,1627,1637,1657,1663,1667,1669,1693,1697,1699,1709,1721,1723,1733,1741,1747,1753,1759,1777,1783,1787,1789,1801,1811,1823,1831,1847,1861,1867,1871,1873,1877,1879,1889,1901,1907,1913,1931,1933,1949,1951,1973,1979,1987,1993,1997,1999,2003,2011,2017,2027,2029,2039,2053,2063,2069,2081,2083,2087,2089,2099,2111,2113,2129,2131,2137,2141,2143,2153,2161,2179,2203,2207,2213,2221,2237,2239,2243,2251,2267,2269,2273,2281,2287,2293,2297,2309,2311,2333,2339,2341,2347,2351,2357,2371,2377,2381,2383,2389,2393,2399,2411,2417,2423,2437,2441,2447,2459,2467,2473,2477,2503,2521,2531,2539,2543,2549,2551,2557,2579,2591,2593,2609,2617,2621,2633,2647,2657,2659,2663,2671,2677,2683,2687,2689,2693,2699,2707,2711,2713,2719,2729,2731,2741,2749,2753,2767,2777,2789,2791,2797,2801,2803,2819,2833,2837,2843,2851,2857,2861,2879,2887,2897,2903,2909,2917,2927,2939,2953,2957,2963,2969,2971,2999
};
struct MyNumber
{
long long t;
long long p;
};
int main() {
int q;
cin >> q;
for (int i = 0; i < q; i++)
{
vector<MyNumber>numberList;
long long num, k;
cin >> num >> k;
for (const auto& it : primeList)
{
if (num == 1) break;
// it 可以整除 num
if (num % it == 0)
{
long long ti = 0;
while (true)
{
num /= it;
ti++;
if (num % it != 0) break;
}
if (ti >= k)
{
MyNumber tt{ ti,it };
numberList.push_back(tt);
}
}
}
// 输出
if (numberList.size() == 0) // 没有剩余项简化后的值等于 1
{
cout << 1 << endl;
}
else
{
long long aws = 1;
for (const auto& it : numberList) {
for (int iii = 0; iii < it.t; iii++)
{
aws *= it.p;
}
}
cout << aws << endl;
}
}
return 0;
}
补充
- 对于5000以内的素数，可以先编写一个小程序把5000以内的素数都筛先选出来，然后存储到
primeList中去，对应代码如下所示：
#include<iostream>
#include<vector>
#include <cmath>
using namespace std;
int main() {
vector<int>primeArr;
for (int i = 2; i < 3000; i++)
{
int flag = 0;
for (int j = 2; j <= sqrt(i); j++)
{
if (i % j == 0)
{
flag = 1;
break;
}
}
if (flag == 0)
{
primeArr.push_back(i);
}
}
for (auto& it : primeArr)
{
cout << it << ",";
}
return 0;
}
