[BZOJ2326] [HNOI2011] 數學作業 (矩陣乘法)

Description

Input

Output

Sample Input

Sample Output

HINT

Source

Solution

　　遞推式長這樣：$f[n]=f[n-1]*10^k+n$

　　對于每一段位數個數相同的$n$（如$10\sim99,100\sim999,23333\sim66666,1018701389\sim2147483647$），$k$是個定值

　　然后就可以開心地分段矩陣乘法了，剩下的自己推吧

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int mod;
struct mat
{
    ll a[4][4];
    int n, m;
 
    mat()
    {
        memset(a, 0, sizeof(a));
        n = 0, m = 0;
    }
 
    mat(int x, int y)
    {
        memset(a, 0, sizeof(a));
        n = x, m = y;
    }
 
    mat operator* (const mat &rhs) const
    {
        mat ans;
        ans.n = n, ans.m = rhs.m;
        for(int i = 1; i <= n; ++i)
            for(int j = 1; j <= rhs.m; ++j)
                for(int k = 1; 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 = 1; i <= n; ++i)
            ans.a[i][i] = 1;
        for(; rhs; rhs >>= 1, b = b * b)
            if(rhs & 1) ans = ans * b;
        return ans;
    }
};
 
int main()
{
    ll n, c;
    mat ans(1, 3), b(3, 3);
    scanf("%lld%d", &n, &mod);
    ans.a[1][3] = 1;
    for(int i = 1; i <= 3; ++i)
        for(int j = 1; j <= i; ++j)
            b.a[i][j] = 1;
    for(ll i = 10; ; i *= 10)
    {
        b.a[1][1] = i % mod;
        if(i <= n) c = i / 10 * 9;
        else c = n - i / 10 + 1;
        ans = ans * (b ^ c);
        if(i > n) break;
    }
    printf("%lld\n", ans.a[1][1]);
    return 0;
}

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