SGU108 Self-numbers 2

SGU108 Self-numbers 2

题目大意

定义d(n)等于n的数位和加n
对于一个数v，若不存在n使得d(n) = v，则称v是self-number
输入N，求在区间[1, N]上有多少个self-number，并输出其中指定下标的K个数

算法思路

直接筛选，这里有两个小技巧：

• 使用bitset代替bool数组以减少内存占用
• 先对s[]数组排序，利用类似two pointer的方法记录答案

时间复杂度: O(Nlog10N)

代码

/** * Copyright (c) 2015 Authors. All rights reserved. * * FileName: 108.cpp * Author: Beiyu Li <[email protected]> * Date: 2015-05-22 */
#include <bits/stdc++.h>

using namespace std;

#define rep(i,n) for (int i = 0; i < (n); ++i)
#define For(i,s,t) for (int i = (s); i <= (t); ++i)
#define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)

typedef long long LL;
typedef pair<int, int> Pii;

const int inf = 0x3f3f3f3f;
const LL infLL = 0x3f3f3f3f3f3f3f3fLL;

const int maxv = 10000000 + 70;
const int maxk = 5000 + 5;

int n, k;
Pii s[maxk];
bitset<maxv> vis;

bool cmp(Pii a, Pii b) { return a.second < b.second; }

int main()
{
        scanf("%d%d", &n, &k);
        rep(i,k) scanf("%d", &s[i].first), s[i].second = i;
        sort(s, s + k);
        int tot = 0, now = 0;
        For(i,1,n) {
                if (!vis.test(i)) {
                        ++tot;
                        while (now < k && tot == s[now].first)
                                s[now++].first = i;
                }
                int t = i, v = i;
                while (t) v += t % 10, t /= 10;
                vis.set(v);
        }
        printf("%d\n", tot);
        sort(s, s + k, cmp);
        rep(i,k) printf("%d%c", s[i].first, i < k - 1? ' ': '\n');

        return 0;
}