目录
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^d2a3b5c7d
.
析:预处理出所有形为2a3b5c7d2^a3^b5^c7^d2a3b5c7d
即可, 大概只有5000左右个.然后用二分查找就好。
代码如下:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 | #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; } |
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