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
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define INIT() ios::sync_with_stdio(false);cin.tie(0);cout.tie(0); 5 #define Rep(i,n) for (int i = 0; i < (n); ++i) 6 #define For(i,s,t) for (int i = (s); i <= (t); ++i) 7 #define rFor(i,t,s) for (int i = (t); i >= (s); --i) 8 #define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i) 9 #define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i) 10 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i) 11 #define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i) 12 13 #define pr(x) cout << #x << " = " << x << " " 14 #define prln(x) cout << #x << " = " << x << endl 15 16 #define LOWBIT(x) ((x)&(-x)) 17 18 #define ALL(x) x.begin(),x.end() 19 #define INS(x) inserter(x,x.begin()) 20 #define UNIQUE(x) x.erase(unique(x.begin(), x.end()), x.end()) 21 #define REMOVE(x, c) x.erase(remove(x.begin(), x.end(), c), x.end()); // 删去 x 中所有 c 22 #define TOLOWER(x) transform(x.begin(), x.end(), x.begin(),::tolower); 23 #define TOUPPER(x) transform(x.begin(), x.end(), x.begin(),::toupper); 24 25 #define ms0(a) memset(a,0,sizeof(a)) 26 #define msI(a) memset(a,inf,sizeof(a)) 27 #define msM(a) memset(a,-1,sizeof(a)) 28 29 #define MP make_pair 30 #define PB push_back 31 #define ft first 32 #define sd second 33 34 template<typename T1, typename T2> 35 istream &operator>>(istream &in, pair<T1, T2> &p) { 36 in >> p.first >> p.second; 37 return in; 38 } 39 40 template<typename T> 41 istream &operator>>(istream &in, vector<T> &v) { 42 for (auto &x: v) 43 in >> x; 44 return in; 45 } 46 47 template<typename T1, typename T2> 48 ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) { 49 out << "[" << p.first << ", " << p.second << "]" << "\n"; 50 return out; 51 } 52 53 inline int gc(){ 54 static const int BUF = 1e7; 55 static char buf[BUF], *bg = buf + BUF, *ed = bg; 56 57 if(bg == ed) fread(bg = buf, 1, BUF, stdin); 58 return *bg++; 59 } 60 61 inline int ri(){ 62 int x = 0, f = 1, c = gc(); 63 for(; c<48||c>57; f = c==‘-‘?-1:f, c=gc()); 64 for(; c>47&&c<58; x = x*10 + c - 48, c=gc()); 65 return x*f; 66 } 67 68 template<class T> 69 inline string toString(T x) { 70 ostringstream sout; 71 sout << x; 72 return sout.str(); 73 } 74 75 inline int toInt(string s) { 76 int v; 77 istringstream sin(s); 78 sin >> v; 79 return v; 80 } 81 82 //min <= aim <= max 83 template<typename T> 84 inline bool BETWEEN(const T aim, const T min, const T max) { 85 return min <= aim && aim <= max; 86 } 87 88 typedef long long LL; 89 typedef unsigned long long uLL; 90 typedef pair< double, double > PDD; 91 typedef pair< int, int > PII; 92 typedef pair< int, PII > PIPII; 93 typedef pair< string, int > PSI; 94 typedef pair< int, PSI > PIPSI; 95 typedef set< int > SI; 96 typedef set< PII > SPII; 97 typedef vector< int > VI; 98 typedef vector< double > VD; 99 typedef vector< VI > VVI; 100 typedef vector< SI > VSI; 101 typedef vector< PII > VPII; 102 typedef map< int, int > MII; 103 typedef map< int, string > MIS; 104 typedef map< int, PII > MIPII; 105 typedef map< PII, int > MPIII; 106 typedef map< string, int > MSI; 107 typedef map< string, string > MSS; 108 typedef map< PII, string > MPIIS; 109 typedef map< PII, PII > MPIIPII; 110 typedef multimap< int, int > MMII; 111 typedef multimap< string, int > MMSI; 112 //typedef unordered_map< int, int > uMII; 113 typedef pair< LL, LL > PLL; 114 typedef vector< LL > VL; 115 typedef vector< VL > VVL; 116 typedef priority_queue< int > PQIMax; 117 typedef priority_queue< int, VI, greater< int > > PQIMin; 118 const double EPS = 1e-8; 119 const LL inf = 0x7fffffff; 120 const LL infLL = 0x7fffffffffffffffLL; 121 LL mod = 1e9 + 7; 122 const int maxN = 1e5 + 7; 123 const LL ONE = 1; 124 const LL evenBits = 0xaaaaaaaaaaaaaaaa; 125 const LL oddBits = 0x5555555555555555; 126 127 struct Matrix{ 128 int row, col; 129 LL MOD; 130 VVL mat; 131 132 Matrix(int r, int c, LL p = mod) : row(r), col(c), MOD(p) { 133 mat.assign(r, VL(c, 0)); 134 } 135 Matrix(const Matrix &x, LL p = mod) : MOD(p){ 136 mat = x.mat; 137 row = x.row; 138 col = x.col; 139 } 140 Matrix(const VVL &A, LL p = mod) : MOD(p){ 141 mat = A; 142 row = A.size(); 143 col = A[0].size(); 144 } 145 146 // x * 单位阵 147 inline void E(int x = 1) { 148 assert(row == col); 149 Rep(i, row) mat[i][i] = x; 150 } 151 152 inline VL& operator[] (int x) { 153 assert(x >= 0 && x < row); 154 return mat[x]; 155 } 156 157 inline Matrix operator= (const VVL &x) { 158 row = x.size(); 159 col = x[0].size(); 160 mat = x; 161 return *this; 162 } 163 164 inline Matrix operator+ (const Matrix &x) { 165 assert(row == x.row && col == x.col); 166 Matrix ret(row, col); 167 Rep(i, row) { 168 Rep(j, col) { 169 ret.mat[i][j] = mat[i][j] + x.mat[i][j]; 170 ret.mat[i][j] %= MOD; 171 } 172 } 173 return ret; 174 } 175 176 inline Matrix operator* (const Matrix &x) { 177 assert(col == x.row); 178 Matrix ret(row, x.col); 179 Rep(k, x.col) { 180 Rep(i, row) { 181 if(mat[i][k] == 0) continue; 182 Rep(j, x.col) { 183 ret.mat[i][j] += mat[i][k] * x.mat[k][j]; 184 ret.mat[i][j] %= MOD; 185 } 186 } 187 } 188 return ret; 189 } 190 191 inline Matrix operator*= (const Matrix &x) { return *this = *this * x; } 192 inline Matrix operator+= (const Matrix &x) { return *this = *this + x; } 193 194 inline void print() { 195 Rep(i, row) { 196 Rep(j, col) { 197 cout << mat[i][j] << " "; 198 } 199 cout << endl; 200 } 201 } 202 }; 203 204 // 矩阵快速幂,计算x^y 205 inline Matrix mat_pow_mod(Matrix x, LL y) { 206 Matrix ret(x.row, x.col); 207 ret.E(); 208 while(y){ 209 if(y & 1) ret *= x; 210 x *= x; 211 y >>= 1; 212 } 213 return ret; 214 } 215 216 LL L, A, B, M, ans; 217 218 // 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。 219 // 有返回 true 没有返回 false 220 LL st = 0, bit = 10, l, r; 221 bool getLR() { 222 if(st >= L || A + st * B >= bit) return false; 223 l = st; 224 r = L - 1; 225 226 while(l < r) { 227 LL mid = (l + r) >> 1; 228 if(A + mid * B < bit) l = mid + 1; 229 else r = mid; 230 } 231 232 if(A + r * B >= bit) --r; 233 l = st; 234 st = r + 1; // 下一个起始位置 235 return true; 236 } 237 238 int main(){ 239 //freopen("MyOutput.txt","w",stdout); 240 //freopen("input.txt","r",stdin); 241 INIT(); 242 cin >> L >> A >> B >> mod; 243 244 For(i, 1, 18) { // 枚举位数 245 if(getLR()) { 246 Matrix mat(3, 3); 247 mat[0][0] = bit % mod; 248 mat[0][1] = mat[0][2] = mat[1][2] = mat[2][1] = 0; 249 mat[1][0] = mat[1][1] = mat[2][2] = 1; 250 mat[2][1] = B % mod; 251 252 mat = mat_pow_mod(mat, r - l + 1); 253 254 Matrix ret(1, 3); 255 ret[0][0] = ans; 256 ret[0][1] = (A + l * B) % mod; 257 ret[0][2] = 1; 258 259 ret *= mat; 260 261 ans = ret[0][0]; 262 } 263 bit *= 10; 264 } 265 cout << ans << endl; 266 return 0; 267 }
AtCoder ABC 129F Takahashi's Basics in Education and Learning
原文地址:https://www.cnblogs.com/zaq19970105/p/11100184.html