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  • ACM A problem is easy

    A problem is easy

    时间限制:1000 ms  |  内存限制:65535 KB
    难度:3
     
    描述
    When Teddy was a child , he was always thinking about some simple math problems ,such as “What it’s 1 cup of water plus 1 pile of dough ..” , “100 yuan buy 100 pig” .etc..

    One day Teddy met a old man in his dream , in that dream the man whose name was“RuLai” gave Teddy a problem :

    Given an N , can you calculate how many ways to write N as i * j + i + j (0 < i <= j) ?

    Teddy found the answer when N was less than 10…but if N get bigger , he found it was too difficult for him to solve.
    Well , you clever ACMers ,could you help little Teddy to solve this problem and let him have a good dream ?
     
    输入
    The first line contain a T(T <= 2000) . followed by T lines ,each line contain an integer N (0<=N <= 10^11).
    输出
    For each case, output the number of ways in one line
    样例输入
    2
    1
    3
    样例输出
    0
    1
    #include <iostream>
    #include <vector>
    #include <cmath>
    using namespace std;
    
    int main(){
        int T;
        cin >> T;
        for(int  icase  = 0 ; icase < T; icase ++){
            int n;
            cin >> n;
            n++;
            int res = 0;
            for(int i = 2; i <= (int)sqrt(n); ++ i){
                if(n%i == 0) res ++ ;
            }
            cout<< res<<endl;
        }
    }
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  • 原文地址:https://www.cnblogs.com/xiongqiangcs/p/3647191.html
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