[BZOJ 2326][HNOI 2011] 数学作业

题目大意

求将从 \(1\)\(n\) 的正整数顺序连接起来得到的数模 \(m\) 的值。

\(1 \leqslant n \leqslant 10^{18}\)

\(1 \leqslant m \leqslant 1,000,000,000\)

题目链接

BZOJ 2326

CodeVS 2314

题解

按位数分开算,记 \(f[n]\) 为答案,有(\(k\) 为位数): \[ f[i] = (f[i - 1] \times 10^k + i) mod \; m \] 每一段(位数相同)用矩阵乘法计算。

代码

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#include <cstdio>
#include <cstring>
// #define DBG
struct Matrix {
long long a[3][3];
Matrix(bool unit) {
memset(a, 0, sizeof (a));
if (unit) for (int i = 0; i < 3; i++) a[i][i] = 1;
}
long long &operator()(int i, int j) {
return a[i][j];
}
const long long &operator()(int i, int j) const {
return a[i][j];
}
#ifdef DBG
void print() const {
puts("Matrix is :");
for (int i = 0; i < 3; i++) {
printf("( ");
for (int j = 0; j < 3; j++) printf("%lld ", a[i][j]);
printf(")\n");
}
}
#endif
} trans(false), f(false);
long long n, m;
long long mul(long long a, long long b) {
long long res = 0;
for (; b; b >>= 1, a = (a + a) % m) if (b & 1) res = (res + a) % m;
return res;
}
Matrix operator*(const Matrix &a, const Matrix &b) {
#ifdef DBG
puts("Matrix multiply:");
a.print();
b.print();
#endif
Matrix res(false);
for (int i = 0; i < 3; i++) for (int j = 0; j < 3; j++) for (int k = 0; k < 3; k++) (res(i, j) += mul(a(i, k), b(k, j))) %= m;
#ifdef DBG
res.print();
#endif
return res;
}
Matrix pow(Matrix a, long long n) {
#ifdef DBG
printf("pow(trans, %lld)\n", n);
#endif
Matrix res(true);
for (; n; n >>= 1, a = a * a) if (n & 1) res = res * a;
return res;
}
void initMatrix() {
f(2, 0) = 1;
trans(0, 1) = trans(0, 2) = trans(1, 1) = trans(1, 2) = trans(2, 2) = 1;
}
void calc(long long k, long long last) {
trans(0, 0) = k % m;
#ifdef DBG
printf("calc(%lld, %lld)\n", k, last);
trans.print();
f.print();
#endif
long long n = last - k / 10 + 1;
f = pow(trans, n) * f;
}
int main() {
scanf("%lld %lld", &n, &m);
initMatrix();
long long k = 10;
while (n >= k) {
#ifdef DBG
printf("k = %lld\n", k);
#endif
calc(k, k - 1);
k *= 10;
}
calc(k, n);
printf("%lld\n", f(0, 0));
return 0;
}