zoukankan      html  css  js  c++  java
  • Uva10290

    Problem H

    {sum+=i++} to Reach N

    Input: standard input

    Output:  standard output

    Memory Limit: 32 MB

    All the positive numbers can be expressed as a sum of one, two or more consecutive positive integers. For example 9 can be expressed in three such ways, 2+3+44+5 or 9. Given an integer less than (9*10^14+1) or (9E14 + 1) or (9*1014 +1) you will have to determine in how many ways that number can be expressed as summation of consecutive numbers.

     

    Input

    The input file contains less than 1100 lines of input. Each line contains a single integer N  (0<=N<= 9E14). Input is terminated by end of file.

     

    Output

    For each line of input produce one line of output. This line contains an integer which tells in how many ways N can be expressed as summation of consecutive integers.

    Sample Input

    9

    11

    12

     

    Sample Output

    3

    2

    2


    题意:问你N能够由多少种方案:连续的x个整数相加和为N

    思路:转化为求奇因数,分解质因数后求排列数
    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <vector>
    #include <string>
    #include <algorithm>
    #include <queue>
    using namespace std;
    const int maxn = 1e7+10;
    typedef long long ll;
    
    bool isPrime[maxn];
    vector<int> prime,cnt;
    ll n;
    
    void getPrime(){
        memset(isPrime,1,sizeof isPrime);
        for(int i = 2; i < maxn; i++){
            if(isPrime[i]){
                prime.push_back(i);
                for(int j = i+i; j < maxn; j+=i){
                    isPrime[j] = 0;
                }
            }
        }
    }
    
    void getDigit(){
    
        while(n%2==0) n/=2;
       // cout<<n<<endl;
        for(int i = 1; i < prime.size()&&n >= prime[i]; i++){
            if(n%prime[i]==0){
                int t = 0;
                while(n%prime[i]==0){
                    n /= prime[i];
                    t++;
                }
                cnt.push_back(t);
            }
        }
        if(n!=1) cnt.push_back(1);
    }
    
    int main(){
    
        getPrime();
        while(~scanf("%lld",&n)){
            cnt.clear();
            getDigit();
            ll ans = 1;
            for(int i = 0; i < cnt.size(); i++) ans *= (cnt[i]+1);
            cout<<ans<<endl;
    
        }
        return 0;
    }
    


  • 相关阅读:
    9.1、PHP 中的函数
    7.2.2、二维数组中包含自定义键数组的打印
    Windows 8 VHD 概述与使用
    8.2、磁盘、目录和文件计算
    7.2.6、随机取出数组中的某个下标
    7.2.3、数组排序
    7.2.7、数组指针的操作
    CentOS6 下 JDK7 + jBoss AS 7 环境搭建
    How to iterate HashMap using JSTL forEach loop
    windows 8 非内置系统管理员获得完整权限的方法
  • 原文地址:https://www.cnblogs.com/mengfanrong/p/3946491.html
Copyright © 2011-2022 走看看