[Hdoj] 4983 goffi and GCD

The meaning of the question is very clear, that is, finding the different logarithm that satisfies gcd (n-a, n) * gcd (n-B, n) = n ^ K (A, B. Obviously gcd (n-a, n) <= N, gcd (n-B, n) <= n. Therefore, when n is not 1 and K is> 2, the condition (a, B) does not exist ). When K = 2, only (n, n) meet the conditions. Therefore, only n = 1 and K = 1 need to be discussed separately:
When n = 1, No matter what value K is, there are only () conditions, and the result is 1;
When K is 1, that is, gcd (n-a, n) * gcd (n-B, n) = N, then gcd (n-a, n) = I, then gcd (n-B, n) = N/I. That is, to obtain the (, b) logarithm and.

 1 #include <cstdio> 2 #include <cmath> 3  4 const int MOD = (1e9+7); 5  6 __int64 getNotDiv(int x) { 7     int i, r = x; 8     __int64 ret = x; 9 10     for (i=2; i*i<=r; ++i) {11         if (x%i == 0) {12             ret -= ret/i;13             while (x%i == 0)14                 x /= i;15         }16     }17     if (x > 1)18         ret -= ret/x;19     return ret;20 }21 22 int main() {23     int n, k;24     int i, j;25     __int64 ans, tmp;26 27     while (scanf("%d %d", &n, &k) != EOF) {28         if (n==1 || k==2)29             printf("1\n");30         else if (k==1) {31             ans = 0;32             for (i=1; i*i<=n; ++i) {33                 if (n%i == 0) {34                     j = n/i;35                     tmp = getNotDiv(i)*getNotDiv(j)%MOD;36                     if (j == i) {37                         ans += tmp;38                     } else {39                         ans += tmp<<1;40                     }41                     ans %= MOD;42                 }43             }44             printf("%I64d\n", ans%MOD);45         } else {46             printf("0\n");47         }48     }49 50     return 0;51 }

 

[Hdoj] 4983 goffi and GCD