标签:最小值 line eps highlight scan include int ack return
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。
#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; }
标签:最小值 line eps highlight scan include int ack return
原文地址:http://www.cnblogs.com/tyty-Somnuspoppy/p/6391047.html