zoukankan      html  css  js  c++  java
  • UVA 294

    Mathematicians love all sorts of odd properties of numbers. For instance, they consider 945 to be an interesting number, since it is the first odd number for which the sum of its divisors is larger than the number itself.

    To help them search for interesting numbers, you are to write a program that scans a range of numbers and determines the number that has the largest number of divisors in the range. Unfortunately, the size of the numbers, and the size of the range is such that a too simple-minded approach may take too much time to run. So make sure that your algorithm is clever enough to cope with the largest possible range in just a few seconds.

    The first line of input specifies the number N of ranges, and each of the N following lines contains a range, consisting of a lower bound Land an upper bound U, where L and U are included in the range. L and U are chosen such that tex2html_wrap_inline42 and tex2html_wrap_inline44 .

    For each range, find the number P which has the largest number of divisors (if several numbers tie for first place, select the lowest), and the number of positive divisors D of P (where P is included as a divisor). Print the text 'Between L and HP has a maximum of Ddivisors.', where LHP, and D are the numbers as defined above.

    3
    1 10
    1000 1000
    999999900 1000000000
    Between 1 and 10, 6 has a maximum of 4 divisors.
    Between 1000 and 1000, 1000 has a maximum of 16 divisors.
    Between 999999900 and 1000000000, 999999924 has a maximum of 192 divisors.



    题意:给你 a,b,让你找出 a,b之间因子最多的数是多少 b-a《=1000 我们枚举就好了
    //meek
    #include<bits/stdc++.h>
    #include <iostream>
    #include <cstdio>
    #include <cmath>
    #include <string>
    #include <cstring>
    #include <algorithm>
    #include<map>
    #include<queue>
    using namespace std ;
    typedef long long ll;
    #define mem(a) memset(a,0,sizeof(a))
    #define pb push_back
    #define fi first
    #define se second
    #define MP make_pair
    
    const int N=550;
    const ll INF = 1ll<<61;
    const int inf = 1000000007;
    const int MOD =   2000000011;
    
    
    ll cal(ll x) {
        ll hav = 0;
       for(ll i = 1; i*i <= x; i++) {
        if(x % i == 0) hav ++;
        if(x % i == 0 && x / i != i) hav++;
       }
       return hav;
    }
    int main() {
        int T;
        ll a,b,ans,tmp;
        scanf("%d",&T);
        while(T--) {
            ans = -1;
            scanf("%lld%lld",&a,&b);
            for(ll i = a; i <= b; i++) {
                ll temp = cal( i );
                if(temp > ans) {
                    ans = temp;
                    tmp = i;
                }
            }
            printf("Between %lld and %lld, %lld has a maximum of %lld divisors.
    ",a,b,tmp,ans);
        }
        return 0;
    }
    代码
  • 相关阅读:
    ubuntu一些记录
    unittest添加测试用例方法
    弹出框处理
    无法连接终端
    Python 断言
    Appium_Python_Api文档
    pycharm快捷键
    appium运行时启动失败
    appium运行时每次默认弹出appiumsetting与unlock重装,关闭这两个步骤的方法
    SpringBoot的jar包引用外部properties文件
  • 原文地址:https://www.cnblogs.com/zxhl/p/5104001.html
Copyright © 2011-2022 走看看