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
