Sumdiv
| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 17387 | Accepted: 4374 |
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).
Source
题意：求A ^ B的所有因子的和；
分析：对A用唯一分解定理分解 A = p1 ^ a1 * p2 ^ a2 * p3 ^ a3 ... * pn ^ an
其中A的因子的个数为 （ 1 + a1) * ( 1 + a2 ) * ( 1 + a3 ) * ... * ( 1 + an)
则A的所有因子的和为 （1 + p1 + p1 ^ 2 + p1 ^ 3 ... + p1 ^ a1) * ( 1 + p2 ^ 1 + p2 ^ 2 + p3 ^ 3 + ... + p2 ^ a2 ) * ... * ( 1 + pn + pn ^ 2 + ... + pn ^ an)
求 a + a ^ 2 + a ^ 3 + ... + a ^ n
如果n为奇数： a + a ^ 2 + ... + a ^ (n / 2 ) + a ^ (n / 2 + 1) + ( a + a ^ 2 + ... + a ^ ( n / 2 ) ) * a ^ ( n / 2 + 1）
如果n为偶数： a + a ^ 2 + ... + a ^ (n / 2 ） + （ a + a ^ 2 + ... + a ^ ( n /. 2) ) * a ^ (n / 2)
1 #include <iostream> 2 #include <cstring> 3 #include <cstdio> 4 #include <algorithm> 5 using namespace std; 6 typedef long long LL; 7 const int Max = 10000; 8 const int Mod = 9901; 9 int prime[Max + 10],flag[Max],cnt; 10 void get_prime() 11 { 12 cnt = 0; 13 memset(flag, 0, sizeof(flag)); 14 for(int i = 2; i <= Max; i++) 15 { 16 if(flag[i] == 0) 17 { 18 flag[i] = 1; 19 prime[++cnt] = i; 20 for(int j = i; j <= Max / i; j++) 21 flag[i * j] = 1; 22 } 23 } 24 } 25 LL pow_mod(LL n, LL k) 26 { 27 LL res = 1; 28 while(k) 29 { 30 if(k & 1) 31 res = res * n % Mod; 32 n = n * n % Mod; 33 k >>= 1; 34 } 35 return res; 36 } 37 LL get_sum(LL n, LL m) 38 { 39 if(m == 0) 40 return 1; 41 if(m & 1) 42 { 43 return get_sum(n, m / 2) *( 1 + pow_mod(n, m / 2 + 1) ) % Mod; 44 } 45 else 46 { 47 return ( get_sum(n, m / 2 - 1) * (1 + pow_mod(n, m / 2 + 1)) % Mod + pow_mod(n, m / 2) ) % Mod; 48 } 49 } 50 int main() 51 { 52 LL a,b; 53 get_prime(); 54 while(scanf("%I64d%I64d", &a, &b) != EOF) 55 { 56 LL ans = 1; 57 if(a == 0 && b) //特殊情况 58 ans = 0; 59 LL m; 60 for(int i = 1; i <= cnt; i++) 61 { 62 if(prime[i] > a) 63 break; 64 m = 0; 65 if(a % prime[i] == 0) 66 { 67 while(a % prime[i] == 0) 68 { 69 a = a / prime[i]; 70 m++; 71 } 72 m = m * b; // m要设成LL，否则这里会溢出 73 ans = ans * get_sum((LL)prime[i], m) % Mod; 74 } 75 } 76 if(a > 1) 77 ans = ans * get_sum(a, b) % Mod; 78 printf("%I64d ", ans); 79 } 80 return 0; 81 }