UVA - 294 Divisors （约数）（数论）

题意：输入两个整数L，U(1<=L<=U<=10,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;

}