1124. 整数幂

using namespace std;
int qpow(int a,int b,int c)
{
        int ans=1;
        while(b){
                if(b&1) ans=(ans*a)%c;
                a=(a*a)%c;
                b=b>>1;
        }
        return ans;
}
int main()
{
        int a,b,c;
        while(1==1){
                cin>>a>>b>>c;
                if(a==0&&b==0&&c==0) break;
                else cout<<qpow(a,b,c)<<endl;
        }
}

#include "bits/stdc++.h"

using namespace std;
using u64 = uint64_t;

int main() {
    for (u64 a, b, c; cin >> a >> b >> c, a;) {
        u64 ans = a % c;
        for (u64 i = 1; i < b; ++i) {
            ans = (ans * a) % c;
        }
        cout << ans << '\n';
    }
}

#include "bits/stdc++.h"

using namespace std;
using u64 = uint64_t;

template <typename UInt> UInt ipow(UInt const kBase, UInt const kExp, UInt const kMod) {
    UInt ret = 1;
    for (UInt i = kExp, base = kBase % kMod; i; i >>= 1) {
        if (i & 1) {
            ret = ret * base % kMod;
        }
        base = base * base % kMod;
    }
    return ret;
} // quick power algorithm for (a^b) % kMod.

int main() {
    for (u64 a, b, c; cin >> a >> b >> c, a;) {
        cout << ipow(a, b, c) << '\n';
    }
}

#include <bits/stdc++.h>
using namespace std;

int power(int a,int b,int p)
{
    int res=1%p;// 最低位为1
    while(b){
        if(b&1)res=res*1ll*a%p;// ci为1，需要乘以a
        a=a*1ll*a%p;// 每次a都平方
        b>>=1;
    }
    return res;
}

int main()
{
    int a,b,c;
    while(1)
    {
        scanf("%d%d%d",&a,&b,&c);
        if(!a&&!b&&!c)break;
        else printf("%d",power(a,b,c));
    }
    return 0;
}