CF1228C. Primes and Multiplication（数学）

Posted 2022-12-09 smallhester

Let’s introduce some definitions that will be needed later.

Let ??????????(??) be the set of prime divisors of ??. For example, ??????????(140)=2,5,7, ??????????(169)=13.

Let ??(??,??) be the maximum possible integer ???? where ?? is an integer such that ?? is divisible by ????. For example:

??(45,3)=9 (45 is divisible by 32=9 but not divisible by 33=27),
??(63,7)=7 (63 is divisible by 71=7 but not divisible by 72=49).
Let ??(??,??) be the product of ??(??,??) for all ?? in ??????????(??). For example:

??(30,70)=??(70,2)⋅??(70,3)⋅??(70,5)=21⋅30⋅51=10,
??(525,63)=??(63,3)⋅??(63,5)⋅??(63,7)=32⋅50⋅71=63.
You have integers ?? and ??. Calculate ??(??,1)⋅??(??,2)⋅…⋅??(??,??)mod(109+7).

Input
The only line contains integers ?? and ?? (2≤??≤109, 1≤??≤1018) — the numbers used in formula.

Output
Print the answer.

Examples
inputCopy
10 2
outputCopy
2
inputCopy
20190929 1605
outputCopy
363165664
inputCopy
947 987654321987654321
outputCopy
593574252
Note
In the first example, ??(10,1)=??(1,2)⋅??(1,5)=1, ??(10,2)=??(2,2)⋅??(2,5)=2.

In the second example, actual value of formula is approximately 1.597⋅10171. Make sure you print the answer modulo (109+7).

In the third example, be careful about overflow issue.

思路：求出x的素因子，求其在1-n中所有数的贡献

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<stack>
#include<cstdlib>
#include<queue>
#include<set>
#include<string.h>
#include<vector>
#include<deque>
#include<map>
using namespace std;
#define INF 0x3f3f3f3f
#define eps 1e-10
#define bug printf("*********\n")
#define debug(x) cout<<#x"=["<<x<<"]" <<endl
typedef long long LL;
typedef long long ll;
const int MAXN = 150000 + 5;
const int mod = 1e9 + 7;
 
vector<LL>v;
LL qpow(LL a,LL b) 
    LL res = 1;
    while(b) 
        if(b & 1)
            res = res * a % mod;
        a = a * a % mod;
        b >>= 1;
    
    return res;

 
 
void primeFactor(LL n)
    LL tmp = n;
    if(n % 2 == 0) 
        v.push_back(2);
        while (n % 2 == 0) 
            n /= 2;
        
    
    for(LL i = 3; i * i <= tmp; i += 2)
        if(n % i == 0) 
            v.push_back(i);
        
        while(n % i == 0)
            n /= i;
        
    
    if(n > 2)
        v.push_back(n);

int main()

    LL x,n;
    cin >> x;
    cin >> n;
    primeFactor(x);
    LL ans = 1;
    for(int i = 0; i < v.size(); i++) 
        LL tt = 0,nn = n;
        while(nn > 0) 
            nn /= v[i];
            tt += nn;
        
        ans = (ans % mod * qpow(v[i],tt)) % mod;
    
 
    cout << ans << endl;
 

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