題目鏈接

題目描述

Consider digits strings with length n, how many different strings have the sum of digits are multiple of 4?

輸入描述:

There are several test cases end with EOF. For each test case, there is an integer in one line: n, the length of the digits string. (1≤n≤1,000,000,000).

輸出描述:

For each case, output the number of digits strings with length n have the sum of digits are multiple of 4 in one line. The numbers maybe very large, you should output the result modular 2019.

輸入

1
2
3
4

輸出

3
25
249
479

思路

長度為1有四種情況：

%4 == 0 有3個

%4 == 1 有3個

%4 == 2 有2個

%4 == 3 有2個

計算長度為N的分為四種情況：
令長度為N-1模4為{0， 1， 2， 3}的數量為T0, T1, T3, T4

%4 == 0
(%4=0) * T0 + (%4=1) * T3 + (%4=2) * T2 + (%4=3) * T1

%4 == 1
(%4=0) * T1 + (%4=1) * T0 + (%4=2) * T3 + (%4=3) * T2

%4 == 2

(%4=0) * T2 + (%4=1) * T1 + (%4=2) * T0 + (%4=3) * T3

%4 == 3
(%4=0) * T3 + (%4=1) * T2 + (%4=2) * T1 + (%4=3) * T0

根據這個思路可以用矩陣快速冪加快一下

#include <bits/stdc++.h> #define LL long long #define P pair<int, int> #define lowbit(x) (x & -x) #define mem(a, b) memset(a, b, sizeof(a)) #define REP(i, n) for (int i = 1; i <= (n); ++i) #define rep(i, n) for (int i = 0; i < (n); ++i) #define N 100005 using namespace std;void mul(LL x[], LL y[]) {LL t[4];t[0] = x[0] * y[0] + x[1] * y[3] + x[3] * y[1] + x[2] * y[2];t[1] = x[1] * y[0] + x[0] * y[1] + x[2] * y[3] + x[3] * y[2];t[2] = x[0] * y[2] + x[2] * y[0] + x[1] * y[1] + x[3] * y[3];t[3] = x[0] * y[3] + x[3] * y[0] + x[1] * y[2] + x[2] * y[1];rep (i, 4) x[i] = t[i] % 2019; }LL quick(LL x) {LL b[4] = {3, 3, 2, 2};LL a[4] = {3, 3, 2, 2};while (x) {if (x & 1) mul(a, b);mul(b, b);x >>= 1;}return a[0]; }int main() { #ifndef ONLINE_JUDGEfreopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout); #endifLL n;while (scanf("%lld", &n) != EOF) {LL ans = quick(n-1);printf("%d\n", ans % 2019);}return 0; }

總結

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