AtCoder ABC 129F Takahashi's Basics in Education and Learning

题目链接：https://atcoder.jp/contests/abc129/tasks/abc129_f

题目大意

　　给定一个长度为 L ，首项为 A，公差为 B 的等差数列 S，将这 L 个数拼起来，记作 N，求 N % M。

分析

　　设 bit(i) 为第 i 项所需要进行的十进制位移。

　　则 $N = S_0 * 10^{bit(0)} + S_1 * 10^{bit(1)} + \dots + S_{L - 1} * 10^{bit(L - 1)}$。

　　一项一项地算是肯定要超时的，不过注意到等差数列的每一项都小于 10¹⁸ ，因此很多项的长度是相等的，也就是说有很多 bit(i) 也是等差数列。

　　于是我们可以按照位数给等差数列分组，最多可分 18 组。

　　举个例子，在区间 [L, R] 上，每一项长度都为 k。

　　记这个区间上所表示的数为 A(k)，A(k) 的每一项设为 $a_i, (L \leq i \leq R)$，则 $a_i = S_i * 10^{bit(i)}, A(k) = \sum_{i = L}^{R} a_i$。

　　不难看出$A(k) = 10^{bit(L)} * \sum_{i = L}^{R} (S_i * 10^{(R - i) * k})$，只要处理后一部分即可。

　　设 $ret(j) = \sum_{i = L}^{j} (S_i * 10^{(j - i) * k}), (L \leq j \leq R)$。

　　则有 $ret(j + 1) = ret(j) * 10^k + S_{j + 1}$，不妨设 ret(L - 1) = 0。

　　于是我们可以构造如下系数矩阵 X：

$$
X = \begin{bmatrix}
10^k & 0 & 0 \\
1 & 1 & 0 \\
0 & B & 1 \\
\end{bmatrix}
$$

　　和如下矩阵 RET(j)：

$$
RET(j) = \begin{bmatrix}
ret(j) & S_{j + 1} & 1
\end{bmatrix} 
$$

　　于是有：

$$
RET(j) = RET(j - 1) * X \\
RET(j) = RET(L - 1) * X^{j - L + 1}
$$

　　如此，通过矩阵快速幂，长度为 k 的一组值很快就被算出来了，然后每一组都分别算一下再加起来即可。

　　PS：在实际实现过程中组与组之间是可以合并的，并不需要单独算出来每一组的余数，详细实现请看代码。

代码如下

 #include <bits/stdc++.h>
 using namespace std;

 #define INIT() ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
 #define Rep(i,n) for (int i = 0; i < (n); ++i)
 #define For(i,s,t) for (int i = (s); i <= (t); ++i)
 #define rFor(i,t,s) for (int i = (t); i >= (s); --i)
 #define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i)
 #define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i)
 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
 #define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)

 #define pr(x) cout << #x << " = " << x << "  "
 #define prln(x) cout << #x << " = " << x << endl

 #define LOWBIT(x) ((x)&(-x))

 #define ALL(x) x.begin(),x.end()
 #define INS(x) inserter(x,x.begin())
 #define UNIQUE(x) x.erase(unique(x.begin(), x.end()), x.end())
 #define REMOVE(x, c) x.erase(remove(x.begin(), x.end(), c), x.end()); // 删去 x 中所有 c
 #define TOLOWER(x) transform(x.begin(), x.end(), x.begin(),::tolower);
 #define TOUPPER(x) transform(x.begin(), x.end(), x.begin(),::toupper);

 #define ms0(a) memset(a,0,sizeof(a))
 #define msI(a) memset(a,inf,sizeof(a))
 #define msM(a) memset(a,-1,sizeof(a))

 #define MP make_pair
 #define PB push_back
 #define ft first
 #define sd second

 template<typename T1, typename T2>
 istream &operator>>(istream &in, pair<T1, T2> &p) {
     in >> p.first >> p.second;
     return in;
 }

 template<typename T>
 istream &operator>>(istream &in, vector<T> &v) {
     for (auto &x: v)
         in >> x;
     return in;
 }

 template<typename T1, typename T2>
 ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) {
     out << "[" << p.first << ", " << p.second << "]" << "\n";
     return out;
 }

 inline int gc(){
     static const int BUF = 1e7;
     static char buf[BUF], *bg = buf + BUF, *ed = bg;

     if(bg == ed) fread(bg = buf, , BUF, stdin);
     return *bg++;
 } 

 inline int ri(){
     int x = , f = , c = gc();
     for(; c<||c>; f = c=='-'?-:f, c=gc());
     for(; c>&&c<; x = x* + c - , c=gc());
     return x*f;
 }

 template<class T>
 inline string toString(T x) {
     ostringstream sout;
     sout << x;
     return sout.str();
 }

 inline int toInt(string s) {
     int v;
     istringstream sin(s);
     sin >> v;
     return v;
 }

 //min <= aim <= max
 template<typename T>
 inline bool BETWEEN(const T aim, const T min, const T max) {
     return min <= aim && aim <= max;
 }

 typedef long long LL;
 typedef unsigned long long uLL;
 typedef pair< double, double > PDD;
 typedef pair< int, int > PII;
 typedef pair< int, PII > PIPII;
 typedef pair< string, int > PSI;
 typedef pair< int, PSI > PIPSI;
 typedef set< int > SI;
 typedef set< PII > SPII;
 typedef vector< int > VI;
 typedef vector< double > VD;
 typedef vector< VI > VVI;
 typedef vector< SI > VSI;
 typedef vector< PII > VPII;
 typedef map< int, int > MII;
 typedef map< int, string > MIS;
 typedef map< int, PII > MIPII;
 typedef map< PII, int > MPIII;
 typedef map< string, int > MSI;
 typedef map< string, string > MSS;
 typedef map< PII, string > MPIIS;
 typedef map< PII, PII > MPIIPII;
 typedef multimap< int, int > MMII;
 typedef multimap< string, int > MMSI;
 //typedef unordered_map< int, int > uMII;
 typedef pair< LL, LL > PLL;
 typedef vector< LL > VL;
 typedef vector< VL > VVL;
 typedef priority_queue< int > PQIMax;
 typedef priority_queue< int, VI, greater< int > > PQIMin;
 const double EPS = 1e-;
 const LL inf = 0x7fffffff;
 const LL infLL = 0x7fffffffffffffffLL;
 LL mod = 1e9 + ;
 const int maxN = 1e5 + ;
 const LL ONE = ;
 const LL evenBits = 0xaaaaaaaaaaaaaaaa;
 const LL oddBits = 0x5555555555555555;

 struct Matrix{
     int row, col;
     LL MOD;
     VVL mat;

     Matrix(int r, int c, LL p = mod) : row(r), col(c), MOD(p) {
         mat.assign(r, VL(c, ));
     }
     Matrix(const Matrix &x, LL p = mod) : MOD(p){
         mat = x.mat;
         row = x.row;
         col = x.col;
     }
     Matrix(const VVL &A, LL p = mod) : MOD(p){
         mat = A;
         row = A.size();
         col = A[].size();
     }

     // x * 单位阵
     inline void E(int x = ) {
         assert(row == col);
         Rep(i, row) mat[i][i] = x;
     }

     inline VL& operator[] (int x) {
         assert(x >=  && x < row);
         return mat[x];
     }

     inline Matrix operator= (const VVL &x) {
         row = x.size();
         col = x[].size();
         mat = x;
         return *this;
     }

     inline Matrix operator+ (const Matrix &x) {
         assert(row == x.row && col == x.col);
         Matrix ret(row, col);
         Rep(i, row) {
             Rep(j, col) {
                 ret.mat[i][j] = mat[i][j] + x.mat[i][j];
                 ret.mat[i][j] %= MOD;
             }
         }
         return ret;
     }

     inline Matrix operator* (const Matrix &x) {
         assert(col == x.row);
         Matrix ret(row, x.col);
         Rep(k, x.col) {
             Rep(i, row) {
                 if(mat[i][k] == ) continue;
                 Rep(j, x.col) {
                     ret.mat[i][j] += mat[i][k] * x.mat[k][j];
                     ret.mat[i][j] %= MOD;
                 }
             }
         }
         return ret;
     }

     inline Matrix operator*= (const Matrix &x) { return *this = *this * x; }
     inline Matrix operator+= (const Matrix &x) { return *this = *this + x; }

     inline void print() {
         Rep(i, row) {
             Rep(j, col) {
                 cout << mat[i][j] << " ";
             }
             cout << endl;
         }
     }
 };

 // 矩阵快速幂，计算x^y
 inline Matrix mat_pow_mod(Matrix x, LL y) {
     Matrix ret(x.row, x.col);
     ret.E();
     while(y){
         if(y & ) ret *= x;
         x *= x;
         y >>= ;
     }
     return ret;
 }

 LL L, A, B, M, ans; 

 // 从数列第 st 项开始，查找区间 [L, R]，使得区间内的所有数都小于 bit 大于等于 bit/10。
 // 有返回 true 没有返回 false
 LL st = , bit = , l, r;
 bool getLR() {
     if(st >= L || A + st * B >= bit) return false;
     l = st;
     r = L - ;

     while(l < r) {
         LL mid = (l + r) >> ;
         if(A + mid * B < bit) l = mid + ;
         else r = mid;
     }

     if(A + r * B >= bit) --r;
     l = st;
     st = r + ; // 下一个起始位置
     return true;
 }

 int main(){
     //freopen("MyOutput.txt","w",stdout);
     //freopen("input.txt","r",stdin);
     INIT();
     cin >> L >> A >> B >> mod;

     For(i, , ) { // 枚举位数
         if(getLR()) {
             Matrix mat(, );
             mat[][] = bit % mod;
             mat[][] = mat[][] = mat[][] = mat[][] = ;
             mat[][] = mat[][] = mat[][] = ;
             mat[][] = B % mod;

             mat = mat_pow_mod(mat, r - l + );

             Matrix ret(, );
             ret[][] = ans;
             ret[][] = (A + l * B) % mod;
             ret[][] = ;

             ret *= mat;

             ans = ret[][];
         }
         bit *= ;
     }
     cout << ans << endl;
     return ;
 }