bzoj 4002: [JLOI2015]有意义的字符串

　　这个题。。。

　　

 #include <bits/stdc++.h>
 #define rep(i, a, b) for (int i = a; i <= b; i++)
 #define drep(i, a, b) for (int i = a; i >= b; i--)
 #define REP(i, a, b) for (int i = a; i < b; i++)
 #define mp make_pair
 #define pb push_back
 #define clr(x) memset(x, 0, sizeof(x))
 #define xx first #define yy second
 using namespace std;
 typedef pair<int, int> pii;
 typedef unsigned long long ll;
 const int inf = 0x3f3f3f3f;
 const ll INF = 0x3f3f3f3f3f3f3f3fll;
 //************************************************

 const ll mod = 7528443412579576937LL;
 ll mul(ll base, ll num) {
     ll ret();
     while (num) {
         if (num & ) ret = (ret + base) % mod;
         base = (base + base) % mod;
         num >>= ;
     }
     return ret;
 }
 struct matrix {
     ll s[][]; matrix () { clr(s); }
     matrix operator * (const matrix &B) const {
         matrix A = *this, C;
         rep(i, , ) rep(j, , ) rep(k, , )
             C.s[i][j] = (C.s[i][j] + mul(A.s[i][k], B.s[k][j]) % mod) % mod;
         return C;
     }
 } ori;

 matrix POW(matrix base, ll num) {
     matrix ret = ori;
     while (num) {
         if (num & ) ret = ret * base;
         base = base * base;
         num >>= ;
     }
     return ret;
 }

 int main() {
     ll b, d, n; scanf("%llu%llu%llu", &b, &d, &n);
     ll a = b, c = (d - b * b) >> ;
     ori.s[][] = ori.s[][] = ;
     matrix base; base.s[][] = a, base.s[][] = c, base.s[][] = ;
     matrix ans;
     ans.s[][] = , ans.s[][] = b;
     ans = ans * POW(base, n);
     if (d != b * b && n %  == ) ans.s[][]--;
     printf("%llu\n", ans.s[][]);
 }