[AHOI2002]网络传输

这道题根据题意，易知k的幂与p的二进制形式有关系，然后再一波高精度即可。（这里我用$n、k$代替了$k、p$）

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>

 using namespace std;

 #define re register
 #define rep(i, a, b) for (re int i = a; i <= b; ++i)
 #define repd(i, a, b) for (re int i = a; i >= b; --i)
 #define maxx(a, b) a = max(a, b);
 #define minn(a, b) a = min(a, b);
 #define LL long long
 #define inf (1 << 30)

 inline int read() {
     int w = , f = ; char c = getchar();
     while (!isdigit(c)) f = c == '-' ? - : f, c = getchar();
     while (isdigit(c)) w = (w << ) + (w << ) + (c ^ ''), c = getchar();
     return w * f;
 }

 const int maxl = , base = ;

 int n, k;

 int a[maxl], sum[maxl];

 void time() {
     int x = ;
     rep(i, , a[]) a[i] = a[i] * n + x, x = a[i] / base, a[i] %= base;
     if (x) a[++a[]] = x;
 }

 void add() {
     int x = ; maxx(sum[], a[]);
     rep(i, , sum[]) sum[i] += a[i] + x, x = sum[i] / base, sum[i] %= base;
     if (x) sum[++sum[]] = x;
 }

 int main() {
     freopen("ques.in", "r", stdin);
     freopen("ques.out", "w", stdout);

     n = read(), k = read();
     a[] = a[] = sum[] = ;

     while (k) {
         if (k & ) add();
         k >>= ;
         time();
     }

     repd(i, sum[], )
         if (i == sum[]) printf("%d", sum[i]); else printf("%04d", sum[i]);

     return ;
 }