hdu2035 people love a^b (fast power to take the mold)

Topic Link:hdu 2035 people to see people love a^b

Very early time to do a problem, today think to turn him out, write an article to do not know the fast power of the students to do a popular science (please allow me to blow a bit). The power operation can be calculated efficiently by the fast power. If we use loops to calculate, then the time complexity is O (n), using the fast power of the word only with O (log n). Do not underestimate such a little, if a problem requires a number of power operations, it may be because of this little change and time-out.

Quick Power Introduction:

We always say fast power, so where is he going? If we solve 2^k. It can be represented as

X^n = ((x2) 2 .....)

As long as the K-squared operation is possible, we can think of, first, n is represented as 2 power-side second and

n = 2^k1 + 2^k2 + 2^k3 ....

's Got

X^n = x^ (2^K1) x^ (2^K2) x^ (2^K3) ....

So fast power is so fast. You can simulate it manually with a pen and paper.

Example: x^22 = x^16 x^4 x^2

Quick-Power templates:

typedef long Long LL; Note that this is not necessarily a long long sometimes int also line ll Mod_pow (ll X, ll N, ll MoD) {    ll res = 1;    while (n > 0) {         if (N & 1) res = res * x% MoD;    N&1 actually here and n%2 expression is a meaning        x = x * x% mod;        n >>= 1;                 N >>= 1 This and n/=2 expression is a meaning    }    return res;

Don't look at the arithmetic of children's shoes, well back to see, a lot of places are using this thing

Recursive version of:

typedef long Long Ll;ll Mod_pow (ll X, ll N, ll MoD) {    if (n = = 0) return 1;    ll res = MOD_POW (x * x% mod, N/2, mod);    if (n & 1) res = res * x% MoD;    return res;}

The following code is attached:

#include <stdio.h>int mod_pow (int x, int n,int mod) {      //fast power    int res = 1;    while (n > 0) {        if (N & 1) res = res * x% MoD;        x = x * x% mod;        n >>= 1;    }    return res;} int main () {     int m,n;    while (scanf ("%d%d", &m,&n), n| | m)        printf ("%d\n", Mod_pow (m,n,1000));    return 0;}

(If there is an error, please correct it, reproduced in the source)



hdu2035 people love a^b (fast power to take the mold)