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HDU 2837 calculation

2013-05-16 17:55 246 查看
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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