HDU 2837 calculation
2013-05-16 17:55
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容易证明 命题 若A^X>M,则 X>=phi(M) 的反例只有 A=2,X=2,M=6 而此特例中 2^1%6=2 2^2%6=4 2^3%6= 2, 当X=1时已经开始循环,故此特例不必特判
#include <utility> #include <algorithm> #include <string> #include <cstring> #include <cstdio> #include <iostream> #include <iomanip> #include <set> #include <vector> #include <cmath> #include <queue> #include <bitset> #include <map> #include <iterator> using namespace std; #define clr(a,v) memset(a,v,sizeof(a)) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 const int INF = 0x7f7f7f7f; const int maxn = 100009; const double pi = acos(-1.0); const double eps = 1e-10; const int mod = 1000000007; typedef long long LL; typedef pair<int, int> pii; typedef vector<int> VI; typedef vector<VI> VVI; typedef vector<VVI> VVVI; bitset<maxn + 100> vis; int p[maxn], size; LL POW(LL a, LL b, int mod) { LL res = 1; while (b) { if (b & 1) res = res * a % mod; a = a * a % mod; b >>= 1; } return res; } void init() { int i, j; vis[0] = vis[1] = 1; for (i = 2; i <= 325; ++i) if (!vis[i]) for (j = i * i; j <= maxn; vis[j] = 1, j += i) ; for (i = 2; i <= maxn; ++i) if (!vis[i]) p[size++] = i; } int phi(int n) { int i, res = n; for (i = 0; 1ll * p[i] * p[i] <= n && i < size; ++i) { if (n % p[i] == 0) { res = res - res / p[i]; n /= p[i]; while (n % p[i] == 0) n /= p[i]; } } if (n > 1) res = res - res / n; return res; } bool cmp(LL a, int k, LL m) { LL res = 1; for (int i = 0; i < k && res < m; ++i) res *= a; return res >= m; } LL dfs(int n, int m) { if (!n) { return 1 % m; } int mod = phi(m); int d = n % 10; int t = n / 10; LL pv = dfs(t, mod); LL res = POW(d, pv, m); res += cmp(d, pv, m) * m; return res; } int main() { ios::sync_with_stdio(false); init(); int T,n,m; cin>>T; while(T--) { cin>>n>>m; cout<<dfs(n,m)%m<<endl; } return 0; }
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