poj 3243 poj 2417 hdu 2815（解高次同余方程的应用）

高次同余方程用babystep_giantstep算法来解，具体原理学不懂，学会了怎么用。

poj 3243：

题意：

已知x，z，k，求同余方程    x ^ y mod z = k   中的y值。

代码：

#include <iostream>#include <cstdio>#include <cstdlib>#include <algorithm>#include <cstring>#include <cmath>#include <stack>#include <vector>#include <queue>#include <map>#include <climits>#include <cassert>#define LL long longusing namespace std;const int inf = 0x3f3f3f3f;const double eps = 1e-8;const double pi = 4 * atan(1.0);const double ee = exp(1.0);const int HASH = 65536;struct hashMap{    int head[HASH], Size;    int next[HASH];    LL state[HASH], val[HASH];    void init()    {        memset(head, -1, sizeof(head));        Size = 0;    }    void Insert(LL st, LL sv)    {        LL h = st % HASH;        for (int p = head[h]; p != -1; p = next[p])            if (state[p] == st)                return;        state[Size] = st;        val[Size] = sv;        next[Size] = head[h];        head[h] = Size++;    }    LL Find(LL st)    {        LL h = st % HASH;        for (int p = head[h]; p != -1; p = next[p])            if (state[p] == st)                return val[p];        return -1;    }} hashmap;LL gcd(LL a, LL b){    return b == 0 ? a : gcd(b, a % b);}void exgcd(LL a, LL b, LL& d, LL& x, LL& y){    if (b == 0)    {        d = a;        x = 1;        y = 0;    }    else    {        exgcd(b, a % b, d, y, x);        y -= a / b * x;    }}LL inv(LL a, LL n){    LL x, y, d;    exgcd(a, n, d, x, y);    return (x + n) % n;}LL pow_mod(LL a, LL n, LL mod){    if (n == 0)        return 1;    LL x = pow_mod(a, n >> 1, mod);    LL res = x * x % mod;    if (n % 2)        res = a * res % mod;    return res;}LL babystep_giantstep(LL a, LL b, LL n){    for (LL i = 0, e = 1; i <= 100; i++)    {        if (e == b)            return i;        e = (e * a) % n;    }    LL k = 1, cnt = 0;    while (1)    {        LL t = gcd(a, n);        if (t == 1)            break;        if (b % t != 0)            return -1;        n /= t;        b /= t;        k = (k * a / t) % n;        cnt++;    }    hashmap.init();    hashmap.Insert(1, 0);    LL e = 1, m = (LL)(sqrt(n + 0.5));    for (int i = 1; i <= m; i++)    {        e = (e * a) % n;        hashmap.Insert(e, i);    }    LL p = inv(pow_mod(a, m, n), n), v = inv(k, n);    for (int i = 0; i <= m; i++)    {        LL t = hashmap.Find((b * v) % n);        if (t != -1)            return i * m + t + cnt;        v = (v * p) % n;    }    return -1;}int main(){#ifdef LOCAL    freopen("in.txt", "r", stdin);#endif // LOCAL    LL x, z, k;    while (scanf("%lld%lld%lld", &x, &z, &k) == 3)    {        if (!x && !z && !k)            break;        LL ans = babystep_giantstep(x % z, k % z, z);        if (ans == -1)            printf("No Solution\n");        else            printf("%lld\n", ans);    }    return 0;}

poj 2417：

同理；

代码：

#include <iostream>#include <cstdio>#include <cstdlib>#include <algorithm>#include <cstring>#include <cmath>#include <stack>#include <vector>#include <queue>#include <map>#include <climits>#include <cassert>#define LL long longusing namespace std;const int inf = 0x3f3f3f3f;const double eps = 1e-8;const double pi = 4 * atan(1.0);const double ee = exp(1.0);const int HASH = 65536;struct hashMap{    int head[HASH], Size;    int next[HASH];    LL state[HASH], val[HASH];    void init()    {        memset(head, -1, sizeof(head));        Size = 0;    }    void Insert(LL st, LL sv)    {        LL h = st % HASH;        for (int p = head[h]; p != -1; p = next[p])            if (state[p] == st)                return;        state[Size] = st;        val[Size] = sv;        next[Size] = head[h];        head[h] = Size++;    }    LL Find(LL st)    {        LL h = st % HASH;        for (int p = head[h]; p != -1; p = next[p])            if (state[p] == st)                return val[p];        return -1;    }} hashmap;LL gcd(LL a, LL b){    return b == 0 ? a : gcd(b, a % b);}void exgcd(LL a, LL b, LL& d, LL& x, LL& y){    if (b == 0)    {        d = a;        x = 1;        y = 0;    }    else    {        exgcd(b, a % b, d, y, x);        y -= a / b * x;    }}LL inv(LL a, LL n){    LL x, y, d;    exgcd(a, n, d, x, y);    return (x + n) % n;}LL pow_mod(LL a, LL n, LL mod){    if (n == 0)        return 1;    LL x = pow_mod(a, n >> 1, mod);    LL res = x * x % mod;    if (n % 2)        res = a * res % mod;    return res;}LL babystep_giantstep(LL a, LL b, LL n){    for (LL i = 0, e = 1; i <= 100; i++)    {        if (e == b)            return i;        e = (e * a) % n;    }    LL k = 1, cnt = 0;    while (1)    {        LL t = gcd(a, n);        if (t == 1)            break;        if (b % t != 0)            return -1;        n /= t;        b /= t;        k = (k * a / t) % n;        cnt++;    }    hashmap.init();    hashmap.Insert(1, 0);    LL e = 1, m = (LL)(sqrt(n + 0.5));    for (int i = 1; i <= m; i++)    {        e = (e * a) % n;        hashmap.Insert(e, i);    }    LL p = inv(pow_mod(a, m, n), n), v = inv(k, n);    for (int i = 0; i <= m; i++)    {        LL t = hashmap.Find((b * v) % n);        if (t != -1)            return i * m + t + cnt;        v = (v * p) % n;    }    return -1;}int main(){#ifdef LOCAL    freopen("in.txt", "r", stdin);#endif // LOCAL    LL p, b, n;    while (scanf("%lld%lld%lld", &p, &b, &n) == 3)    {        LL ans = babystep_giantstep(b % p, n % p, p);        if (ans == -1)            printf("no solution\n");        else            printf("%lld\n", ans);    }    return 0;}

hdu 2815：

题意：

给一棵树，每个节点有k个儿子，求这棵树的最小深度d，使得这个深度的节点数对p取模的结果为n。

解析：

一棵满树深度为d时的节点数为 k ^ d 。

所以题目转化为 k ^ d mod p = n.

另外，如果n > p，显然无解。

代码：

#include <iostream>#include <cstdio>#include <cstdlib>#include <algorithm>#include <cstring>#include <cmath>#include <stack>#include <vector>#include <queue>#include <map>#include <climits>#include <cassert>#define LL long longusing namespace std;const int inf = 0x3f3f3f3f;const double eps = 1e-8;const double pi = 4 * atan(1.0);const double ee = exp(1.0);const int HASH = 65536;struct hashMap{    int head[HASH], Size;    int next[HASH];    LL state[HASH], val[HASH];    void init()    {        memset(head, -1, sizeof(head));        Size = 0;    }    void Insert(LL st, LL sv)    {        LL h = st % HASH;        for (int p = head[h]; p != -1; p = next[p])            if (state[p] == st)                return;        state[Size] = st;        val[Size] = sv;        next[Size] = head[h];        head[h] = Size++;    }    LL Find(LL st)    {        LL h = st % HASH;        for (int p = head[h]; p != -1; p = next[p])            if (state[p] == st)                return val[p];        return -1;    }} hashmap;LL gcd(LL a, LL b){    return b == 0 ? a : gcd(b, a % b);}void exgcd(LL a, LL b, LL& d, LL& x, LL& y){    if (b == 0)    {        d = a;        x = 1;        y = 0;    }    else    {        exgcd(b, a % b, d, y, x);        y -= a / b * x;    }}LL inv(LL a, LL n){    LL x, y, d;    exgcd(a, n, d, x, y);    return (x + n) % n;}LL pow_mod(LL a, LL n, LL mod){    if (n == 0)        return 1;    LL x = pow_mod(a, n >> 1, mod);    LL res = x * x % mod;    if (n % 2)        res = a * res % mod;    return res;}LL babystep_giantstep(LL a, LL b, LL n){    for (LL i = 0, e = 1; i <= 100; i++)    {        if (e == b)            return i;        e = (e * a) % n;    }    LL k = 1, cnt = 0;    while (1)    {        LL t = gcd(a, n);        if (t == 1)            break;        if (b % t != 0)            return -1;        n /= t;        b /= t;        k = (k * a / t) % n;        cnt++;    }    hashmap.init();    hashmap.Insert(1, 0);    LL e = 1, m = (LL)(sqrt(n + 0.5));    for (int i = 1; i <= m; i++)    {        e = (e * a) % n;        hashmap.Insert(e, i);    }    LL p = inv(pow_mod(a, m, n), n), v = inv(k, n);    for (int i = 0; i <= m; i++)    {        LL t = hashmap.Find((b * v) % n);        if (t != -1)            return i * m + t + cnt;        v = (v * p) % n;    }    return -1;}int main(){#ifdef LOCAL    freopen("in.txt", "r", stdin);#endif // LOCAL    LL k, p, n;    while (scanf("%I64d%I64d%I64d", &k, &p, &n) == 3)    {        if (p < n)            printf("Orz,I can’t find D!\n");        else        {            LL ans = babystep_giantstep(k % p, n % p, p);            if (ans == -1)                printf("Orz,I can’t find D!\n");            else                printf("%I64d\n", ans);        }    }    return 0;}