fzu 1759 Super A^B mod C 大数幂取模

/*
 * FZU1759.cpp
 *
 *  Created on: 2011-10-11
 *      Author: bjfuwangzhu
 */
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>
#define LL long long
#define nnum 1000005
#define nmax 31625
int flag[nmax], prime[nmax];
int plen;
void mkprime() {
    int i, j;
    memset(flag, -1, sizeof(flag));
    for (i = 2, plen = 0; i < nmax; i++) {
        if (flag[i]) {
            prime[plen++] = i;
        }
        for (j = 0; (j < plen) && (i * prime[j] < nmax); j++) {
            flag[i * prime[j]] = 0;
            if (i % prime[j] == 0) {
                break;
            }
        }
    }
}
int getPhi(int n) {
    int i, te, phi;
    te = (int) sqrt(n * 1.0);
    for (i = 0, phi = n; (i < plen) && (prime[i] <= te); i++) {
        if (n % prime[i] == 0) {
            phi = phi / prime[i] * (prime[i] - 1);
            while (n % prime[i] == 0) {
                n /= prime[i];
            }
        }
    }
    if (n > 1) {
        phi = phi / n * (n - 1);
    }
    return phi;
}
int cmpBigNum(int p, char *ch) {
    int i, len;
    LL res;
    len = strlen(ch);
    for (i = 0, res = 0; i < len; i++) {
        res = (res * 10 + (ch[i] - '0'));
        if (res > p) {
            return 1;
        }
    }
    return 0;
}
int getModBigNum(int p, char *ch) {
    int i, len;
    LL res;
    len = strlen(ch);
    for (i = 0, res = 0; i < len; i++) {
        res = (res * 10 + (ch[i] - '0')) % p;
    }
    return (int) res;
}
int modular_exp(int a, int b, int c) {
    LL res, temp;
    res = 1 % c, temp = a % c;
    while (b) {
        if (b & 1) {
            res = res * temp % c;
        }
        temp = temp * temp % c;
        b >>= 1;
    }
    return (int) res;
}
void solve(int a, int c, char *ch) {
    int phi, res, b;
    phi = getPhi(c);
    if (cmpBigNum(phi, ch)) {
        b = getModBigNum(phi, ch) + phi;
    } else {
        b = atoi(ch);
    }
    res = modular_exp(a, b, c);
    printf("%d\n", res);
}
int main() {
#ifndef ONLINE_JUDGE
    freopen("data.in", "r", stdin);
#endif
    int a, c;
    char cha[nnum], chb[nnum];
    mkprime();
    while (~scanf("%s %s %d", cha, chb, &c)) {
        a = getModBigNum(c, cha);
        solve(a, c, chb);
    }
    return 0;
}

 1000MS

/*
 * FZU1759.cpp
 *
 *  Created on: 2010-9-17
 *      Author: bjfuwangzhu
 */
#include<iostream>
#include<cmath>
#include<string.h>
using namespace std;
const int Val = 50001;
int prime[Val];
void mkPrime() {
    int i, j;
    memset(prime, 1, sizeof(prime));
    for (i = 2; i < Val; i++) {
        if (prime[i]) {
            for (j = i + i; j < Val; j += i)
                prime[j] = 0;
        }
    }
    for (i = 2, j = 0; i < Val; i++)
        if (prime[i])
            prime[j++] = i;
}
long long numPhi(long long n) {
    int i, te;
    long long phi;
    te = static_cast<int>(sqrt(n * 1.0));phi
    = n;
    for (i = 0; prime[i] <= te; i++) {
        if (n % prime[i] == 0) {
            phi = phi / prime[i] * (prime[i] - 1);
            while (n % prime[i] == 0)
                n /= prime[i];
        }
    }
    if (n > 1)
        phi = phi / n * (n - 1);
    return phi;
}
int numPow(long long n, long long m, long long r) {
    long long p = 1, a = static_cast<long long>(n);
    while (m) {
        if (m & 1)
            p = p * a % r;
        m >>= 1;
        a = a * a % r;
    }
    return static_cast<int>(p);
}
int getSum(char *b) {
    int sum = 0;
    int len = strlen(b);
    for (int i = 0; i < len; i++)
        sum = sum * 10 + b[i] - '0';
    return sum;
}
long long Solve(char *b, long long phi) {
    long long sum = 0;
    int len = strlen(b);
    for (int i = 0; i < len; i++) {
        sum = sum * 10 + b[i] - '0';
        sum %= phi;
    }
    return sum;
}
bool numCmp(char *b, long long phi) {
    long long sum = 0;
    int len = strlen(b);
    for (int i = 0; i < len; i++) {
        sum = sum * 10 + b[i] - '0';
        if (sum > phi)
            return true;
    }
    return false;
}
int main() {
    mkPrime();
    char r[1000010];
    long long a, b, phi, num;
    while (cin >> a >> r >> b) {
        phi = numPhi(b);
        a %= b;
        if (numCmp(r, phi))
            num = Solve(r, phi) + phi;
        else
            num = getSum(r);
        cout << numPow(a, num, b) << endl;
    }
    return 0;
}

125MS

/*
 * FZU1759.cpp
 *
 *  Created on: 2011-10-11
 *      Author: bjfuwangzhu
 */
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>
#define LL long long
#define nnum 1000005
#define nmax 31625
int flag[nmax], prime[nmax];
int plen;
void mkprime() {
    int i, j;
    memset(flag, -1, sizeof(flag));
    for (i = 2, plen = 0; i < nmax; i++) {
        if (flag[i]) {
            prime[plen++] = i;
        }
        for (j = 0; (j < plen) && (i * prime[j] < nmax); j++) {
            flag[i * prime[j]] = 0;
            if (i % prime[j] == 0) {
                break;
            }
        }
    }
}
int getPhi(int n) {
    int i, te, phi;
    te = (int) sqrt(n * 1.0);
    for (i = 0, phi = n; (i < plen) && (prime[i] <= te); i++) {
        if (n % prime[i] == 0) {
            phi = phi / prime[i] * (prime[i] - 1);
            while (n % prime[i] == 0) {
                n /= prime[i];
            }
        }
    }
    if (n > 1) {
        phi = phi / n * (n - 1);
    }
    return phi;
}
int cmpCphi(int p, char *ch) {
    int i, len;
    LL res;
    len = strlen(ch);
    for (i = 0, res = 0; i < len; i++) {
        res = (res * 10 + (ch[i] - '0'));
        if (res > p) {
            return 1;
        }
    }
    return 0;
}
int getCP(int p, char *ch) {
    int i, len;
    LL res;
    len = strlen(ch);
    for (i = 0, res = 0; i < len; i++) {
        res = (res * 10 + (ch[i] - '0')) % p;
    }
    return (int) res;
}
int modular_exp(int a, int b, int c) {
    LL res, temp;
    res = 1 % c, temp = a % c;
    while (b) {
        if (b & 1) {
            res = res * temp % c;
        }
        temp = temp * temp % c;
        b >>= 1;
    }
    return (int) res;
}
void solve(int a, int c, char *ch) {
    int phi, res, b;
    phi = getPhi(c);
    if (cmpCphi(phi, ch)) {
        b = getCP(phi, ch) + phi;
    } else {
        b = atoi(ch);
    }
    res = modular_exp(a, b, c);
    printf("%d\n", res);
}
int main() {
#ifndef ONLINE_JUDGE
    freopen("data.in", "r", stdin);
#endif
    int a, c;
    char ch[nnum];
    mkprime();
    while (~scanf("%d %s %d", &a, ch, &c)) {
        solve(a % c, c, ch);
    }
    return 0;
}

当A 、B很大时