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。

#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;
}