题目

You are given a positive integer n > 1. Consider all the different prime divisors of n. Each of them is included in the expansion n into prime factors in some degree. Required to find among the indicators of these powers is minimal.

数据范围

T ≤ 50000

It is guaranteed that each of them does not exceed 10^18.

题解

题意

给你一个数字x,求将其按照唯一分解定理分解后最低次幂是多少

真·题解

考场上想了很多,但最后没想到这是个思维题

我们首先用不超过的素数分解,然后我们用表示剩下的部分,这也就是说表示剩下的所有的素数的因数都比大,剩下的数只能是,,或者是(和都是素数)(因为如果再大就会超过M了)

代码

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
bool isprime[100100];
int prime[10000]={0,2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197,199,211,223,227,229,233,239,241,251,257,263,269,271,277,281,283,293,307,311,313,317,331,337,347,349,353,359,367,373,379,383,389,397,401,409,419,421,431,433,439,443,449,457,461,463,467,479,487,491,499,503,509,521,523,541,547,557,563,569,571,577,587,593,599,601,607,613,617,619,631,641,643,647,653,659,661,673,677,683,691,701,709,719,727,733,739,743,751,757,761,769,773,787,797,809,811,821,823,827,829,839,853,857,859,863,877,881,883,887,907,911,919,929,937,941,947,953,967,971,977,983,991,997,1009,1013,1019,1021,1031,1033,1039,1049,1051,1061,1063,1069,1087,1091,1093,1097,1103,1109,1117,1123,1129,1151,1153,1163,1171,1181,1187,1193,1201,1213,1217,1223,1229,1231,1237,1249,1259,1277,1279,1283,1289,1291,1297,1301,1303,1307,1319,1321,1327,1361,1367,1373,1381,1399,1409,1423,1427,1429,1433,1439,1447,1451,1453,1459,1471,1481,1483,1487,1489,1493,1499,1511,1523,1531,1543,1549,1553,1559,1567,1571,1579,1583,1597,1601,1607,1609,1613,1619,1621,1627,1637,1657,1663,1667,1669,1693,1697,1699,1709,1721,1723,1733,1741,1747,1753,1759,1777,1783,1787,1789,1801,1811,1823,1831,1847,1861,1867,1871,1873,1877,1879,1889,1901,1907,1913,1931,1933,1949,1951,1973,1979,1987,1993,1997,1999,2003,2011,2017,2027,2029,2039,2053,2063,2069,2081,2083,2087,2089,2099,2111,2113,2129,2131,2137,2141,2143,2153,2161,2179,2203,2207,2213,2221,2237,2239,2243,2251,2267,2269,2273,2281,2287,2293,2297,2309,2311,2333,2339,2341,2347,2351,2357,2371,2377,2381,2383,2389,2393,2399,2411,2417,2423,2437,2441,2447,2459,2467,2473,2477,2503,2521,2531,2539,2543,2549,2551,2557,2579,2591,2593,2609,2617,2621,2633,2647,2657,2659,2663,2671,2677,2683,2687,2689,2693,2699,2707,2711,2713,2719,2729,2731,2741,2749,2753,2767,2777,2789,2791,2797,2801,2803,2819,2833,2837,2843,2851,2857,2861,2879,2887,2897,2903,2909,2917,2927,2939,2953,2957,2963,2969,2971,2999,3001,3011,3019,3023,3037,3041,3049,3061,3067,3079,3083,3089,3109,3119,3121,3137,3163,3167,3169,3181,3187,3191,3203,3209,3217,3221,3229,3251,3253,3257,3259,3271,3299,3301,3307,3313,3319,3323,3329,3331,3343,3347,3359,3361,3371,3373,3389,3391,3407,3413,3433,3449,3457,3461,3463,3467,3469,3491,3499,3511,3517,3527,3529,3533,3539,3541,3547,3557,3559,3571,3581,3583,3593,3607,3613,3617,3623,3631,3637,3643,3659,3671,3673,3677,3691,3697,3701,3709,3719,3727,3733,3739,3761,3767,3769,3779,3793,3797,3803,3821,3823,3833,3847,3851,3853,3863,3877,3881,3889,3907,3911,3917,3919,3923,3929,3931,3943,3947,3967,3989,4001,4003,4007,4013,4019,4021,4027,4049,4051,4057,4073,4079,4091,4093,4099,4111,4127,4129,4133,4139,4153,4157,4159,4177,4201,4211,4217,4219,4229,4231,4241,4243,4253,4259,4261,4271,4273,4283,4289,4297,4327,4337,4339,4349,4357,4363,4373,4391,4397,4409,4421,4423,4441,4447,4451,4457,4463,4481,4483,4493,4507,4513,4517,4519,4523,4547,4549,4561,4567,4583,4591,4597,4603,4621,4637,4639,4643,4649,4651,4657,4663,4673,4679,4691,4703,4721,4723,4729,4733,4751,4759,4783,4787,4789,4793,4799,4801,4813,4817,4831,4861,4871,4877,4889,4903,4909,4919,4931,4933,4937,4943,4951,4957,4967,4969,4973,4987,4993,4999,5003,5009,5011,5021,5023,5039,5051,5059,5077};
int main()
{
   //freopen("a.txt","r",stdin);freopen("a.out","w",stdout);
   ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
   int n;
   cin>>n;
   int i=1;

   while(n--)
  {
       int k=0,maxx=0x3f3f;
       LL num;cin>>num;
       for(int i=1;i<=551;i++){if(num%prime[i]==0){while(num%prime[i]==0){num/=prime[i];k++;}maxx=min(k,maxx);k=0;}}
       LL temp=pow(num,1.0/5);
       if(temp<2){cout<<maxx<<endl;continue;}
       if(pow(temp,1.0/4)-(int)pow(temp,1.0/4)==0){cout<<4<<endl;continue;}
       if(pow(temp,1.0/3)-(int)pow(temp,1.0/3)==0){cout<<3<<endl;continue;}
       if(sqrt(temp)-(int)sqrt(temp)==0){cout<<2<<endl;continue;}
       cout<<1<<endl;
  }

   return 0;
}

Humble Halten