Sumdiv
Time Limit: 1000MS | Memory Limit: 30000K | |
Total Submissions: 16466 | Accepted: 4101 |
Description
Consider two natural numbers A and B. Let S be the sum of all natural divisors of A^B. Determine S modulo 9901 (the rest of the division of S by 9901).
Input
The only line contains the two natural numbers A and B, (0 <= A,B <= 50000000)separated by blanks.
Output
The only line of the output will contain S modulo 9901.
Sample Input
2 3
Sample Output
15
Hint
2^3 = 8.
The natural divisors of 8 are: 1,2,4,8. Their sum is 15.
15 modulo 9901 is 15 (that should be output).
The natural divisors of 8 are: 1,2,4,8. Their sum is 15.
15 modulo 9901 is 15 (that should be output).
要求的是A^B的所有因子的和之后再mod 9901的值。
因为一个数A能够表示成多个素数的幂相乘的形式。即A=(a1^n1)*(a2^n2)*(a3^n3)...(am^nm)。所以这个题就是要求
(1+a1+a1^2+...a1^n1)*(1+a2+a2^2+...a2^n2)*(1+a3+a3^2+...a3^n2)*...(1+am+am^2+...am^nm) mod 9901。
对于每一个(1+a1+a1^2+...a1^n1) mod 9901
等于 (a1^(n1+1)-1)/(a1-1) mod 9901,这里用到逆元的知识:a/b mod c = (a mod (b*c))/ b
所以就等于(a1^(n1+1)-1)mod (9901*(a1-1)) / (a1-1)。
至于前面的a1^(n1+1),快速幂。
代码:
#include <iostream> #include <algorithm> #include <cmath> #include <vector> #include <string> #include <cstring> #pragma warning(disable:4996) using namespace std; #define M 9901 long long p[100008]; int prime[100008]; void isprime() { int cnt = 0, i, j; memset(prime, 0, sizeof(prime)); for (i = 2; i < 100008; i++) { if (prime[i] == 0) { p[++cnt] = i; for (j = 2 * i; j <100008;j=j+i) { prime[j] = 1; } } } } long long getresult(long long A,long long n,long long k) { long long b = 1; while (n > 0) { if (n & 1) { b = (b*A)%k; } n = n >> 1; A = (A*A)%k; } return b; } void solve(long long A, long long B) { int i; long long ans = 1; for (i = 1; p[i] * p[i] <= A; i++) { if (A%p[i] == 0) { int num = 0; while (A%p[i] == 0) { num++; A = A / p[i]; } long long m = (p[i] - 1) * 9901; ans *= (getresult(p[i], num*B + 1, m) + m - 1) / (p[i] - 1); ans %= 9901; } } if (A > 1) { long long m = 9901 * (A - 1); ans *= (getresult(A, B + 1, m) + m - 1) / (A - 1); ans %= 9901; } cout << ans << endl; } int main() { long long A, B; isprime(); while (scanf("%lld%lld", &A, &B) != EOF) { solve(A, B); } return 0; }
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