解题思路：p[i][j]表示说长度为i，且最后一个数字为j的数字占长度为i的数组的百分比。
#include#include int n, k; double p[105][15]; void solve() { memset(p, 0, sizeof(p)); for (int i = 0; i <= k; i++) p[1][i] = 1.0 / (k + 1); for (int i = 2; i <= n; i++) { for (int j = 0; j <= k; j++) { p[i][j] = p[i - 1][j]; if (j) p[i][j] += p[i - 1][j - 1]; if (j != k) p[i][j] += p[i - 1][j + 1]; p[i][j] /= (k + 1); } } } int main () { while (scanf("%d%d", &k, &n) == 2) { solve(); double ans = 0; for (int i = 0; i <= k; i++) ans += p[n][i]; printf("%.5lf\n", ans * 100); } return 0; }