CF1228C. Primes and Multiplication(数学)

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
#include
#include
#include
#include
#include
#include
#include
#include<set>
#include<string.h>
#include
#include
#include
using namespace std;
#define INF 0x3f3f3f3f
#define eps 1e-10
#define bug printf("*********\n")
#define debug(x) cout<<#x"=["<
typedef long long LL;
typedef long long ll;
const int MAXN = 150000 + 5;
const int mod = 1e9 + 7;
 
vectorv;
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;
 
}
View Code

 

转载于:https://www.cnblogs.com/smallhester/p/11614893.html

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