[HNOI2011]数学作业 矩阵快速幂 BZOJ 2326
题目描述
小 C 数学成绩优异,于是老师给小 C 留了一道非常难的数学作业题:
给定正整数 NNN 和 MMM ,要求计算Concatenate(1..N) Concatenate (1 .. N) Concatenate(1..N) ModModMod MMM 的值,其中 Concatenate(1..N) Concatenate (1 .. N) Concatenate(1..N) 是将所有正整数 1,2,…,N1, 2, …, N1,2,…,N 顺序连接起来得到的数。例如,N=13N = 13N=13 , Concatenate(1..N)=12345678910111213Concatenate (1 .. N)=12345678910111213Concatenate(1..N)=12345678910111213 .小C 想了大半天终于意识到这是一道不可能手算出来的题目,于是他只好向你求助,希望你能编写一个程序帮他解决这个问题。
输入输出格式
输入格式:从文件input.txt中读入数据,输入文件只有一行且为用空格隔开的两个正整数N和M,其中30%的数据满足1≤N≤10000001≤N≤10000001≤N≤1000000 ;100%的数据满足1≤N≤10181≤N≤10^{18}1≤N≤1018 且1≤M≤1091≤M≤10^91≤M≤109 .
输出格式:输出文件 output.txt 仅包含一个非负整数,表示 Concatenate(1..N)Concatenate (1 .. N)Concatenate(1..N) ModModMod MMM 的值。
输入输出样例
输入样例#1:
复制
13 13
输出样例#1: 复制
4
范围1e18,logn算法;
递推方程: f[ n ]= f[ n-1 ] * 10^( len(n) )+ n;
直接递推显然不行;
换成矩阵加速递推;
( fn,n,1 )= ( f(n-1),n-1,1 )( 10^(len) , 0 , 0
1 , 1 , 0
1 , 1 , 1 );
这里的 len 指的是 n 的位数;
那么我们每次枚举其位数进行计算即可;
这里我的快速幂参考了别人的封装;
#include<iostream> #include<cstdio> #include<algorithm> #include<cstdlib> #include<cstring> #include<string> #include<cmath> #include<map> #include<set> #include<vector> #include<queue> #include<bitset> #include<ctime> #include<deque> #include<stack> #include<functional> #include<sstream> //#include<cctype> //#pragma GCC optimize(2) using namespace std; #define maxn 2000005 #define inf 0x7fffffff //#define INF 1e18 #define rdint(x) scanf("%d",&x) #define rdllt(x) scanf("%lld",&x) #define rdult(x) scanf("%lu",&x) #define rdlf(x) scanf("%lf",&x) #define rdstr(x) scanf("%s",x) typedef long long ll; typedef unsigned long long ull; typedef unsigned int U; #define ms(x) memset((x),0,sizeof(x)) const long long int mod = 1e9 + 7; #define Mod 1000000000 #define sq(x) (x)*(x) #define eps 1e-3 typedef pair<int, int> pii; #define pi acos(-1.0) //const int N = 1005; #define REP(i,n) for(int i=0;i<(n);i++) typedef pair<int, int> pii; inline ll rd() { ll x = 0; char c = getchar(); bool f = false; while (!isdigit(c)) { if (c == '-') f = true; c = getchar(); } while (isdigit(c)) { x = (x << 1) + (x << 3) + (c ^ 48); c = getchar(); } return f ? -x : x; } ll gcd(ll a, ll b) { return b == 0 ? a : gcd(b, a%b); } ll sqr(ll x) { return x * x; } /*ll ans; ll exgcd(ll a, ll b, ll &x, ll &y) { if (!b) { x = 1; y = 0; return a; } ans = exgcd(b, a%b, x, y); ll t = x; x = y; y = t - a / b * y; return ans; } */ ll mode; struct mat { ll n, m, a[5][5]; mat(ll n, ll m) { this->n = n; this->m = m; ms(a); } mat(ll n, ll m, char e) { this->n = n; this->m = m; ms(a); for (int i = 1; i <= n; i++)a[i][i] = 1; } ll *operator[](const ll x) { return a[x]; } mat operator*(mat b) { mat c(n, b.m); for (int i = 1; i <= n; i++) { for (int j = 1; j <= b.m; j++) { for (int k = 1; k <= m; k++) c[i][j] = (c[i][j] + a[i][k] % mode*b[k][j] % mode) % mode; } } return c; } void operator*=(mat& b) { *this = *this*b; } mat operator ^(ll b) { mat ans(n, m, 'e'), a = *this; while (b) { if (b % 2)ans = ans * a; a *= a; b >>= 1; } return ans; } }; ll n; int getlen(ll x) { int len = 0; while (x) { len++; x /= 10; } return len; } ll pow10(int x) { ll ans = 1; for (int i = 1; i <= x; i++)ans *= 10; return ans; } int main() { //ios::sync_with_stdio(0); rdllt(n); rdllt(mode); mat A(1, 3); int len = getlen(n); A[1][1] = A[1][2] = A[1][3] = 1; for (int i = 0; i < len - 1; i++) { mat B(3,3); B[1][1] = pow10(i + 1) % mode; B[2][1] = B[2][2] = B[3][1] = B[3][2] = B[3][3] = 1; ll m = (pow10(i + 1) - pow10(i)); A = A * (B ^ (m - (i == 0 ? 1 : 0))); } mat B(3, 3); B[1][1] = pow10(len) % mode; B[2][1] = B[2][2] = B[3][1] = B[3][2] = B[3][3] = 1; ll m = n - pow10(len - 1) + 1; A = A * (B ^ (m - (len - 1 == 0 ? 1 : 0))); cout << (ll)A[1][1] << endl; return 0; }
EPFL - Fighting