Fzu 1752 a ^ B mod C fzu 1650 AB mod C

A * B mod C quick calculation method

17:11:18 | category: classic algorithm | Tag: | large font size, small/medium subscription

Method 1:

Everyone can think of it. Calculate the value of a * B and then calculate the value of a * B mod C. This is the simplest, but there is a drawback: the value of a * B cannot be too large and may overflow.

Method 2:

Review the knowledge of hexadecimal conversion. Binary Conversion to hexadecimal can be added with the weight of 2 (which seems to be described in this way ). For example, 13 = (1101) 2 = 1*2 ^ 3 + 1*2 ^ 2 + 0*2 ^ 1 + 1*2 ^ 0. Similarly, when we calculate a * B, we can also convert B to the formula of adding 2 ^ n. Therefore, we can convert a * B mod C to [A * (2 ^ b0 + 2 ^ B1 + ...... 2 ^ bn)] mod c = [A * 2 ^ b0 + A * 2 ^ B1 + ...... A * 2 ^ bn] mod C. The formula (a + B) mod c = [(a mod c) + (B mod C)] mod C is used for calculation.

Code:

Int MUL (int A, int B, int C)

{

Int result, TMP;

TMP = A % C;

While (B)

{

If (B & 1) // determine whether to add a * 2 ^ n based on the value of 2 binary digits, because a right shift operation is performed on B, therefore, you only need to judge the last result each time.

{

Result + = TMP;

If (result> = C)

Result-= C;

}

TMP <= 1; // calculate the value of a * 2 ^ n.

If (TMP> = C)

TMP-= C;

B <= 1;

}

Return result;

}

 

/*
 * FZU1759.cpp
 *
 *  Created on: 2011-10-11
 *      Author: bjfuwangzhu
*/
#include<stdio.h>
#define LL unsigned long long
LL modular_multi(LL a, LL b, LL c) {
    LL res;
    res = 0;
while (b) {
if (b & 1) {
            res += a;
if (res >= c) {
                res -= c;
            }
        }
        a <<= 1;
if (a >= c) {
            a -= c;
        }
        b >>= 1;
    }
return res;
}
LL modular_exp(LL a, LL b, LL c) {
    LL res, temp;
    res = 1 % c, temp = a % c;
while (b) {
if (b & 1) {
            res = modular_multi(res, temp, c);
        }
        temp = modular_multi(temp, temp, c);
        b >>= 1;
    }
return res;
}
int main() {
#ifndef ONLINE_JUDGE
    freopen("data.in", "r", stdin);
#endif
    LL a, b, c;
while (~scanf("%I64u %I64u %I64u", &a, &b, &c)) {
        printf("%I64u\n", modular_exp(a, b, c));
    }
return 0;
}