HDU-5878-I Count Two Three

ACM模版

描述

题解

预处理出来一些 2a3b5c7d 的数，从小到大存起来，当然预处理的手段就多了去了，反正都是打表，然后呢，在这个序列中二分查找即可了。

代码

#include <iostream>#include <cstdio>#include <algorithm>using namespace std;const int MAXN = 5555;const int MAXM = 4;int bs[MAXN] = {1, 2, 3, 4, 5,                6, 7, 8, 9, 10};int tp[MAXM] = {2, 3, 5, 7};void init(){    int cnt;    for (int i = 10; i < MAXN; i++)    {        cnt = i - 1;        bs[i] = INT_MAX;        for (int j = 0; j < 4; j++)        {            while (bs[cnt - 1] * tp[j] > bs[i - 1])            {                cnt--;            }            bs[i] = min(bs[i], bs[cnt] * tp[j]);        }    }}int n;int main(){    init();    int t;    cin >> t;    while (t--)    {        scanf("%d", &n);        int res = *lower_bound(bs, bs + MAXN, n);        printf("%d\n", res);    }       }