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;
}