Let’s consider K-based numbers, containing exactly N digits. We define a number to be valid if its K-based notation doesn’t contain two successive zeros. For example:
1010230 is a valid 7-digit number;1000198 is not a valid number;0001235 is not a 7-digit number, it is a 4-digit number.
Given two numbers N and K, you are to calculate an amount of valid K based numbers, containing N digits. You may assume that 2 ≤ K ≤ 10; N ≥ 2; N + K ≤ 18. Input The numbers N and K in decimal notation separated by the line break. Output The result in decimal notation. Sample input output
2 10
90
dp[i][j]i位数,第i位j时的合法数目
/*************************************************************************> File Name: TOJ1009.cpp> Author: ALex> Mail: zchao1995@gmail.com> Created Time: 2015年05月14日 星期四 18时55分28秒 ************************************************************************/;const double pi = acos(-1.0);const int inf = 0x3f3f3f3f;const double eps = 1e-15;LL;typedef pair <int, int> PLL;LL dp[20][20];int main() {int k, n;while (~scanf(“%d%d”, &n, &k)) {memset(dp, 0, sizeof(dp));for (int i = 0; i < k; ++i) {dp[1][i] = 1;}for (int i = 2; i <= n; ++i) {for (int j = 1; j < k; ++j) {dp[i][0] += dp[i – 1][j];}for (int j = 1; j < k; ++j) {for (int l = 0; l < k; ++l) {dp[i][j] += dp[i – 1][l];}}}LL ans = 0;for (int i = 1; i < k; ++i) {ans += dp[n][i];}printf(“%lld\n”, ans);}return 0;}
，美好的生命应该充满期待、惊喜和感激
