hdu 5878 I Count Two Three ICPC青岛站网络赛1001

题目：

I Count Two Three

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 246    Accepted Submission(s): 128

Problem Description

I will show you the most popular board game in the Shanghai Ingress Resistance Team.
It all started several months ago.
We found out the home address of the enlightened agent Icount2three and decided to draw him out.
Millions of missiles were detonated, but some of them failed.

After the event, we analysed the laws of failed attacks.
It's interesting that the i-th attacks failed if and only if i can be rewritten as the form of 2a3b5c7d which a,b,c,d are non-negative integers.

At recent dinner parties, we call the integers with the form 2a3b5c7d "I Count Two Three Numbers".
A related board game with a given positive integer n from one agent, asks all participants the smallest "I Count Two Three Number" no smaller than n.

 

Input

The first line of input contains an integer t (1≤t≤500000), the number of test cases. t test cases follow. Each test case provides one integer n (1≤n≤109).

 

Output

For each test case, output one line with only one integer corresponding to the shortest "I Count Two Three Number" no smaller than n.

 

Sample Input

1011113123123412345123456123456712345678123456789

 

Sample Output

11214125125012348123480123480012348000123480000

思路：

打表+快速幂

代码：

#include<iostream>#include<algorithm>//3440k 546ms#include<cmath>#include<cstdio>#include<queue>#include<stack>#include<vector>#include<string>#include<cstring>using namespace std;typedef long long ll;ll mod_pow(ll x, ll n){    ll res = 1;    while (n > 0){        if (n & 1) res = res*x;        x = x*x;        n >>= 1;    }    return res;}int main(){    vector <ll> que;//一维向量     for (int i = 0; i <= 30; ++i){        for (int j = 0; j <= 19; ++j){            for (int k = 0; k <= 13; ++k){                for (int l = 0; l <= 11; ++l){                    que.push_back(mod_pow(2, i)*mod_pow(3, j)*mod_pow(5, k)*mod_pow(7, l));                }            }        }    }    sort(que.begin(), que.end());    int t;    scanf("%d", &t);    while (t--){        int n;        scanf("%d", &n);        int i = 0;        printf("%lld\n", *lower_bound(que.begin(), que.end(), n));//返回第一个>=n 的地址中存储的内容，即容器中最小的>=k的数     }}