sgu 108 Self-numbers II

    这道题难在 hash 上， 求出答案很简单， 关键是我们如何标记， 由于 某个数变换后最多比原数多63 所以我们只需开一个63的bool数组就可以了！

    同时注意一下， 可能会有相同的询问。

    我为了防止给的询问不是有序的，还排了一边序。

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <cmath>
#define N 100
#define M 5010
using namespace std;

struct sss
{
    int num, place;
}ask[M];
int n, K;
int ans[M];
bool pd[N]={0};

bool cmp(sss x, sss y)
{
    if (x.num == y.num) return x.place < y.place;
    else return x.num < y.num;
}

int calc(int now)
{
    int a = 0;
    while (now)
    {
        a += now % 10;
        now /= 10;
    }
    return a;
}

int main()
{
    scanf("%d%d", &n, &K);
    for (int i = 1; i <= K; ++i)
    {
        scanf("%d", &ask[i].num);
        ask[i].place = i;
    }
    sort(ask+1, ask+K+1, cmp);
    int num = 0, nowask = 1;
    int lastnum = 0;
    for (int i = 1; i <= n; ++i)
    {
        if (!pd[i%64])
        {
            num++;
            while (num == ask[nowask].num)
            {
                ans[ask[nowask].place] = i;
                nowask++;
            }
        }
        pd[i%64] = 0;
        if (i % 10 == 0)
        {
            lastnum = calc(i);
            pd[(lastnum+i) % 64] = 1;
        }
        else
        {
            lastnum++;
            pd[(lastnum+i) % 64] = 1;
        }
    }
    printf("%d\n", num);
    for (int i = 1; i < K; ++i)
        printf("%d ", ans[i]);
    printf("%d\n", ans[K]);
    return 0;
}