BestCoder Round #62 (div.2) HDOJ5564 Clarke and digits(dp + 快速幂)

Clarke and digits

Time Limit: 5000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 88    Accepted Submission(s): 44

Problem Description

Clarke is a patient with multiple personality disorder. One day, Clarke turned into a researcher, did a research on digits. 
He wants to know the number of positive integers which have a length in [l,r] and are divisible by 7 and the sum of any adjacent digits can not be k.

 

Input

The first line contains an integer T(1≤T≤5), the number of the test cases. 
Each test case contains three integers l,r,k(1≤l≤r≤109,0≤k≤18).

 

Output

Each test case print a line with a number, the answer modulo 109+7.

 

Sample Input

21 2 52 3 5

 

Sample Output

13125Hint:At the first sample there are 13 number $7,21,28,35,42,49,56,63,70,77,84,91,98$ satisfied.   

 

题目链接:点击打开链接

dp[i][j]表示长度为i取模为7的种类数, 参考数据范围, 直接计算会TLE, 用矩阵快速幂来加速运算, 第一次做矩阵快速幂的题目, 参考了题

解, 1LL防止数据溢出出错, 转态转移方程: z.dp[i][j] = (z.dp[i][j] + 1LL * a.dp[i][k] * b.dp[k][j] % MOD) % MOD

AC代码:

#include "iostream"#include "cstdio"#include "cstring"#include "algorithm"#include "cmath"#include "utility"#include "map"#include "set"#include "vector"using namespace std;typedef long long ll;const int MOD = 1e9 + 7;const int MAXN = 100;struct Matrix{/* data */int row, col, dp[MAXN][MAXN];void init(int r, int c) {row = r;col = c;memset(dp, 0, sizeof(dp));}}x, y, z;int l, r, k;void magic(int k){for(int i = 1; i < 10; ++i)x.dp[i * 7 + i % 7][0]++;for(int i = 0; i < 10; ++i)for(int j = 0; j < 7; ++j) {int u = i * 7 + j;for(int m = 0; m < 10; ++m) if(i + m != k) {int v = m * 7 + (j * 10 + m) % 7;y.dp[v][u]++;}}for(int i = 0; i <= 10; ++i)y.dp[70][i * 7] = 1;}void mul(const Matrix &a, const Matrix &b, Matrix &c){z.init(a.row, b.col);for(int i = 0; i < z.row; ++i)for(int j = 0; j < z.col; ++j)for(int k = 0; k < a.col; ++k)z.dp[i][j] = (z.dp[i][j] + 1LL * a.dp[i][k] * b.dp[k][j] % MOD) % MOD;c = z;}int power_mod(Matrix a, Matrix b, int n){while(n > 0) {if(n & 1) mul(b, a, a);mul(b, b, b);n >>= 1;}return a.dp[70][0];}int main(int argc, char const *argv[]){int t;scanf("%d", &t);while(t--) {x.init(MAXN, 1);y.init(MAXN, MAXN);scanf("%d%d%d",&l, &r, &k);magic(k);int ans = (power_mod(x, y, r) - power_mod(x, y, l - 1) + MOD) % MOD;printf("%d\n", ans);}return 0;}