原题目：

题目大意：18位数里头，有多少个数，对于每个数字0-9，在这18位里面出现均不超过3次

111222333444555666 布星~~
112233445566778899 可以~~
 

解题思路：

动态规划

代码:

ll F[19][3000000];void solve() {ll i, j,k,x,y,z,p,q,u,v;ll N = 18,NN=4;double a, b, c,d;F[0][0] = 1;for (i = 1; i <= N; i++) {for (j = 0; j <= 9; j++) {if (j == 0 && i == N)continue;for (k = 0; k <= (1 << 21); k++) {p = k;u = 0;while (p) {z = (3 & p);u = u + z;p = p >> 2;if (u > i - 1)break;}if (u == i - 1) {p = k;v = 3 & (p >> (2 * j));if (v+1 >= NN)continue;F[i][k + ((ll)1 << (2 * j))] += F[i-1][k];}}}}for (i = 1; i <= (1 << 21); i++) {p = i; u = 0; flag = 1;while (p) {z = (3 & p);u = u + z;if (z > NN) { flag = 0; break; }p = p >> 2;}if (u == N&&flag==1) {printf("%lld %lld\n", i,F[N][i]);ans1 = ans1 + F[N][i];}}printf("%lld\n",ans1);
}