HDU 5878 I Count Two Three (预处理+二分查找)

题意：给出一个整数nnn, 找出一个大于等于nnn的最小整数mmm, 使得mmm可以表示为2a3b5c7d2^a3^b5^c7^d2​a​​3​b​​5​c​​7​d​​.

析：预处理出所有形为2a3b5c7d2^a3^b5^c7^d2​a​​3​b​​5​c​​7​d​​即可, 大概只有5000左右个.然后用二分查找就好。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <cstdio>

#include <string>

#include <cstdlib>

#include <cmath>

#include <iostream>

#include <cstring>

#include <set>

#include <queue>

#include <algorithm>

#include <vector>

#include <map>

#include <cctype>

#include <cmath>

#include <stack>

#define freopenr freopen("in.txt", "r", stdin)

#define freopenw freopen("out.txt", "w", stdout)

using namespace std;

typedef long long LL;

typedef pair<int, int> P;

const int INF = 0x3f3f3f3f;

const double inf = 0x3f3f3f3f3f3f;

const double PI = acos(-1.0);

const double eps = 1e-8;

const int maxn = 1e3 + 5;

const int mod = 1e9 + 7;

const int dr[] = {-1, 0, 1, 0};

const int dc[] = {0, 1, 0, -1};

const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};

int n, m;

const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

inline int Min(int a, int b){ return a < b ? a : b; }

inline int Max(int a, int b){ return a > b ? a : b; }

inline LL Min(LL a, LL b){ return a < b ? a : b; }

inline LL Max(LL a, LL b){ return a > b ? a : b; }

inline bool is_in(int r, int c){

    return r >= 0 && r < n && c >= 0 && c < m;

}

vector<LL> ans;

void init(){

    LL a = 1, b = 1, c = 1, d = 1;

    for(; ; a *= 2){

        if(a > mod)  break;

        b = 1;

        for(; ; b *= 3){

            if(a * b > mod) break;

            c = 1;

            for(; ; c *= 5){

                if(a * b * c > mod)  break;

                d = 1;

                for(; ; d *= 7){

                    LL sum = a * b * c * d;

                    if(sum > mod)  break;

                    ans.push_back(sum);

                }

            }

        }

    }

    sort(ans.begin(), ans.end());

}

int main(){

    init();

    int T; cin >> T;

    while(T--){

        LL n;

        scanf("%I64d", &n);

        printf("%I64d\n", *lower_bound(ans.begin(), ans.end(), n));

    }

    return 0;

}