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HDU 4352 XHXJ's LIS 数位DP + 状压

由LIS的nlogn解法 可以得出最后统计数组中数的个数即为LIS的长度 这样就可以状压了

#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <climits>#include <string>#include <iostream>#include <map>#include <cstdlib>#include <list>#include <set>#include <queue>#include <stack>using namespace std;typedef long long LL;const int maxn = 20;LL a,b,k;int len = 0,lim[maxn];LL f[maxn][1 << 10][11];void getlim(LL n) {    len = 0;    memset(lim,0,sizeof(lim));    while(n) {        lim[len++] = n % 10; n /= 10;    }}int check(int state) {    int cnt = 0;    while(state) {        cnt += state & 1;        state >>= 1;    }    return cnt;}int gao(int state,int num) {    for(int i = num;i <= 9;i++) if(state & (1 << i)) {        state &= ~(1 << i);        break;    }    state |= (1 << num);    return state;}LL dfs(int now,int state,int first,int bound) {    if(now == 0) {        if(check(state) == k) {            return 1;        }        return 0;    }    LL &note = f[now][state][k];    int m = bound ? lim[now - 1] : 9;    if(!bound && first && note != -1) return note;    LL ret = 0;    for(int i = 0;i <= m;i++) {        int ns = gao(state,i);        if(i || first) ret += dfs(now - 1,ns,1,bound && i == m);        else ret += dfs(now - 1,state,0,bound && i == m);    }    if(!bound && first) note = ret;    return ret;}LL solve(LL num) {    getlim(num);    return dfs(len,0,0,1);}int main() {    memset(f,-1,sizeof(f));    int T,kase = 1;    cin >> T;    while(T--) {        cout << "Case #" << kase++ << ":" << " ";        cin >> a >> b >> k;        cout << solve(b) - solve(a - 1) << endl;    }    return 0;}