[BZOJ2326] [HNOI2011] 数学作业 (矩阵乘法)

Description

Input

Output

Sample Input

Sample Output

HINT

Source

Solution

　　递推式长这样：$f[n]=f[n-1]*10^k+n$

　　对于每一段位数个数相同的$n$（如$10\sim99,100\sim999,23333\sim66666,1018701389\sim2147483647$），$k$是个定值

　　然后就可以开心地分段矩阵乘法了，剩下的自己推吧

 #include <bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 int mod;
 struct mat
 {
     ll a[][];
     int n, m;

     mat()
     {
         memset(a, , sizeof(a));
         n = , m = ;
     }

     mat(int x, int y)
     {
         memset(a, , sizeof(a));
         n = x, m = y;
     }

     mat operator* (const mat &rhs) const
     {
         mat ans;
         ans.n = n, ans.m = rhs.m;
         for(int i = ; i <= n; ++i)
             for(int j = ; j <= rhs.m; ++j)
                 for(int k = ; k <= m; ++k)
                     ans.a[i][j] = (ans.a[i][j] + a[i][k] * rhs.a[k][j]) % mod;
         return ans;
     }

     mat operator^ (ll rhs) const
     {
         mat ans(n, n), b = *this;
         for(int i = ; i <= n; ++i)
             ans.a[i][i] = ;
         for(; rhs; rhs >>= , b = b * b)
             if(rhs & ) ans = ans * b;
         return ans;
     }
 };

 int main()
 {
     ll n, c;
     mat ans(, ), b(, );
     scanf("%lld%d", &n, &mod);
     ans.a[][] = ;
     for(int i = ; i <= ; ++i)
         for(int j = ; j <= i; ++j)
             b.a[i][j] = ;
     for(ll i = ; ; i *= )
     {
         b.a[][] = i % mod;
         if(i <= n) c = i /  * ;
         else c = n - i /  + ;
         ans = ans * (b ^ c);
         if(i > n) break;
     }
     printf("%lld\n", ans.a[][]);
     return ;
 }