SGU Self-number2

我们考虑，因为最多n为10^7, 7*9 = 63，也就是说一个数x如果不是self-number，那么他的原数一定在[x-63, x]之间。

由于空间有限，所以我们可以用大小>=64的数组优化（只要能存下前63个数即可）。

另外一个优化，打表处理10^4之内的每位数字之和， 则d(x) = x + sum[x/10000] + sum[x%10000];

还有一个坑点是， 查询数字的值不一定是递增的， 也可能有重复的查询。。。

/*Author :usedrose  */
/*Created Time :2015/7/24 11:09:55*/
/*File Name :2.cpp*/
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <cstdlib>
#include <cstring>
#include <climits>
#include <vector>
#include <string>
#include <ctime>
#include <cmath>
#include <deque>
#include <queue>
#include <stack>
#include <set>
#include <map>
#define INF 0x3f3f3f3f
#define eps 1e-8
#define pi acos(-1.0)
#define MAXN 10110
#define OK cout << "ok" << endl;
#define o(a) cout << #a << " = " << a << endl
#define o1(a,b) cout << #a << " = " << a << "  " << #b << " = " << b << endl
using namespace std;
typedef long long LL;

struct node {
    int num, index, ans;
}a[MAXN];
bool f[100];
int n, m;
int sum[MAXN];

bool cmp1(node a, node b)
{
    return a.index < b.index;
}

bool cmp2(node a, node b)
{
    return a.num < b. num;
}
void init()
{
    for (int i = 1;i <= 10000; ++ i)
    {
        int k = i;
        while (k) {
            sum[i] += k%10;
            k /= 10;
        }
    }
}

int main()
{
    //freopen("data.in","r",stdin);
    //freopen("data.out","w",stdout);
    cin.tie(0);
    ios::sync_with_stdio(false);
    init();

    while (cin >> n >> m) {
        for (int i = 1;i <= m; ++ i) {
            cin >> a[i].index;
            a[i].num = i;
        }
        sort(a+1, a+1+m, cmp1);
        int di = 0, cur = 1;
        memset(f, 1, sizeof(f));
        for (int i = 1;i <= n; ++ i) {
            if (f[i%64]) {
                di++;
                while (a[cur].index == di && cur <= m) {
                    a[cur].ans = i;
                    cur++;
                }
            }
            f[i%64] = 1;
            int nxt = i + sum[i/10000] + sum[i%10000];
            f[nxt%64] = 0;
        }

        cout << di << endl;
        sort(a+1, a+1+m, cmp2);
        cout << a[1].ans;
        for (int i = 2;i <= m; ++ i)
            cout << " " << a[i].ans; 
        cout << endl;
    }
    return 0;
}