hdu2035 A^b (congruence and its basic properties, fast power)

began to use the POW function, 6789^10000 calculate the negative, it should be super, but the internet search for a bit said C + + POW () is to support the Longlong to provide overloaded functions of 51456681, is the result of super???

Later, I found that the loop internal variables with the int ...

And then ... Circular *a, or wa ...

Focus on this: have been riding down, the result will be super, there is a mathematical rule: the result of each time to 1000 to take the remainder, the last three will not change, do not understand a word, fill up the knowledge of number theory (congruence operation and its basic properties)

http://www.matrix67.com/blog/archives/236

Code:

#include <iostream>#include<stdio.h>using namespacestd;intMain () {Long Long intn,m;  while(cin>>n>>m) {if(n==0&&m==0)         Break; Long Long intt=1;  for(Long Long inti =1; i<=m;i++) {T*=N; if(t>= +) T= t% +; } cout<<t<<Endl; }    return 0;}

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Fast power:

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).

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) ...

Quick Power Templates:

typedefLong Longll//Note that this is not always a long long and sometimes int is okll Mod_pow (ll X, ll N, ll MoD) {ll res=1;  while(N >0 ){         if(N &1res = res * x% MoD;//n&1 actually here and n%2 expressed a meaningx = x * x%MoD; N>>=1;//N >>= 1 This and n/=2 expression is a meaning    }    returnRes;}

Recursive version:

Long Long Ll;ll Mod_pow (ll X, ll N, ll MoD) {    if0return1;     2 , MoD);     if 1 ) res = res * x% mod;     return Res;}

Solve this problem with a quick power:

#include <stdio.h>intMod_pow (intXintNintMoD) {//Fast Power    intres =1;  while(N >0 ){        if(N &1) Res = res * x%MoD; X= x * x%MoD; N>>=1; }    returnRes;}intMain () {intM,n;  while(SCANF ("%d%d", &m,&n), n| |m) printf ("%d\n", Mod_pow (M,n, +)); return 0;}

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hdu2035 A^b (congruence and its basic properties, fast power)