题目链接 : http://poj.org/problem?id=1845
题意 : 求∑(x^n的所有因子) mod 9901;
思路 : 如果 x = p1^a1 * p2^a2 * p3^a3...pn^an, (p1, p2, ...都是x的质因子), 那么所求解S = (1 + p1 + p1^2 ...+p1^(a1*n)) * (1 + p2 +....).....
所以其实只要解决一个等比数列求和问题就行了, 坑的是这题给的mod 有点小, 如果用公式 (x ^(n+1) - 1) / (x - 1)去做得话, 需要求逆元, 但是存在逆元需要满足
gcd(a, b) == 1
这一性质。 我是用快速幂来做的。
CODE :
#include <cstdio> #include <cstring> #include <algorithm> using namespace std; typedef __int64 lld; const int mod = 9901; struct mat{ lld a[4][4]; }; void init(mat & a){ for (int i = 1; i <= 3; i++) for (int j = 1; j <= 3; j++) a.a[i][j] = 0; } mat mul(mat a, mat b){ mat ans; for (int i = 1; i <= 3; i++){ for (int j = 1; j <= 3; j++){ ans.a[i][j] = 0; for (int k = 1; k <= 3; k++) ans.a[i][j] = (ans.a[i][j] + (a.a[i][k] % mod) * (b.a[k][j] % mod)) % mod; } } return ans; } void debug(mat & a){ for (int i = 1; i <= 3; i++){ for (int j = 1; j <= 3; j++)printf("%d ", a.a[i][j]); printf("\n"); } } lld calc(lld x, lld n){ mat ans, e; init(e); init(ans); for (int i = 1; i <= 3; i++)ans.a[i][i] = 1; e.a[1][1] = 1; e.a[2][2] = x; e.a[3][1] = 1; e.a[3][3] = x; while (n){ if (n & 1){ ans = mul(ans, e); } e = mul(e, e); n >>= 1; } //debug(ans); return (ans.a[3][1] + ans.a[3][3]) % mod; } const int maxn = 10005; int cnt[maxn], p[maxn]; lld solve(lld x, lld n){ memset(p, 0, sizeof(p)); memset(cnt, 0, sizeof(cnt)); int m = 0; for (int i = 2; i <= mod && x > 1; i++){ if (x % i == 0 && x > 1){ p[++m] = i; while (x % i == 0 && x > 1){ cnt[m]++; x /= i; } } } if (x > 1){ m++; p[m] = x; cnt[m] = 1; } lld res = 1; for (int i = 1; i <= m; i++){ res = (res * calc(p[i], cnt[i] * n)) % mod; } return res; } lld x, n; int main(){ while(scanf("%I64d%I64d", &x, &n) != EOF){ printf("%I64d\n", solve(x, n)); } return 0; }