UVA - 294 Divisors (约数)(数论)
题意:输入两个整数L,U(1<=L<=U<=109,U-L<=10000),统计区间[L,U]的整数中哪一个的正约数最多。如果有多个,输出最小值。
分析:
1、求一个数的约数,相当于分解质因子。
2、例如60 = 2 * 2 * 3 * 5。对于2来说,可选0个2,1个2,2个2,有3种情况,同理对于3,有2种情况,对于5,有2种情况,所以3 * 2 * 2则为60的约数个数。
3、L到U扫一遍,取最大值即可。
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 76 77 78 79 80 81 82 83 84 85 | #pragma comment(linker, "/STACK:102400000, 102400000") #include<cstdio> #include<cstring> #include<cstdlib> #include<cctype> #include<cmath> #include<iostream> #include<sstream> #include<iterator> #include<algorithm> #include<string> #include<vector> #include<set> #include<map> #include<stack> #include<deque> #include<queue> #include<list> #define Min(a, b) ((a < b) ? a : b) #define Max(a, b) ((a < b) ? b : a) const double eps = 1e-8; inline int dcmp( double a, double b) { if ( fabs (a - b) < eps) return 0; return a < b ? -1 : 1; } typedef long long LL; typedef unsigned long long ULL; const int INT_INF = 0x3f3f3f3f; const int INT_M_INF = 0x7f7f7f7f; const LL LL_INF = 0x3f3f3f3f3f3f3f3f; const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f; const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1}; const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1}; const int MOD = 1e9 + 7; const double pi = acos (-1.0); const int MAXN = 35000 + 10; const int MAXT = 10000 + 10; using namespace std; int vis[MAXN]; vector< int > prime; void init(){ for ( int i = 2; i < MAXN; ++i){ if (!vis[i]){ prime.push_back(i); for ( int j = 2 * i; j < MAXN; j += i){ vis[j] = 1; } } } } int cal( int n){ int ans = 1; int len = prime.size(); for ( int i = 0; i < len; ++i){ if (prime[i] > n) break ; if (n % prime[i]) continue ; int cnt = 1; while (n % prime[i] == 0){ ++cnt; n /= prime[i]; } ans *= cnt; } return ans; } int main(){ init(); int T; scanf ( "%d" , &T); while (T--){ int l, r; scanf ( "%d%d" , &l, &r); int ans = 0; int id; for ( int i = l; i <= r; ++i){ int tmp = cal(i); if (tmp > ans){ ans = tmp; id = i; } } printf ( "Between %d and %d, %d has a maximum of %d divisors.\n" , l, r, id, ans); } return 0; } |
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