BZOJ 1008 越狱

1008

思路:

总共有m^n次方种方案,其中相邻两个都不相同的有m * (m - 1) ^ (n - 1)种,两者相减就是答案

代码:

#include<bits/stdc++.h>
using namespace std;
#define LL long long 
#define pb push_back
#define mem(a, b) memset(a, b, sizeof(a))
 
const int mod = 1e5 + 3;
LL q_pow(LL n, LL k) {
    LL ans = 1;
    while (k) {
        if (k & 1) ans = (ans * n) % mod;
        n = (n * n) % mod;
        k >>= 1; 
    }
    return ans;
}
int main() {
    LL m, n;
    scanf("%lld%lld", &m, &n);
    LL tot = q_pow(m % mod, n);
    LL tmp = (m % mod)*q_pow((m - 1) % mod, n - 1);
    tot = ((tot - tmp) % mod + mod) % mod; 
    printf("%lld\n", tot);
    return 0;
}

 

posted @ 2018-04-13 10:40  Wisdom+.+  阅读(70)  评论(0编辑  收藏