HDU 2601 An easy problem 因式分解

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An easy problem

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5726    Accepted Submission(s): 1394

Problem Description

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 ?

 

Input

The first line contain a T(T <= 2000) . followed by T lines ,each line contain an integer N (0<=N <= 10¹⁰).

 

Output

For each case, output the number of ways in one line.

 

Sample Input

213

 

Sample Output

01

 

Author

Teddy

 

Source

HDU 1st “Vegetable-Birds Cup” Programming Open Contest

i*j+i+j=n => (i+1)*(j+1)=n+1 所以只要求出(n+1)/(i+1)求出有多少j符合情况就行了。题目中要求j是大于0的，所以要判断求出都j是否大于0，可是一旦加上这个判断，就会超时。

通过证明，这个判断确实是多余的，因为一旦(n+1)%(i+1)==0，就能求出j+1,因为i>0，所以for循环是从2开始的（因为是（i+1）），而i<=j，所以只需要判断到根号n就可以了，因为到了根号n以后情况就重复了，如2*6与6*2只有一种符合情况。那么如果j只有两种情况了，一种是j==0，一种是j>0，显然我们求的是j>0都情况，那么j=0的时候，也就是(i+1）*（0+1）=n+1，可以求出i==n，而for循环只循环到根号n，所以这种情况不可能发生。说的有点罗嗦了。

不过即使是这样，也跑了2000多ms。另外是不是只要求出n+1所有的因子小于等于根号n+1的个数就是次题都解？

//2390MS228K#include<stdio.h>#include<math.h>int main(){    int t;    scanf("%d",&t);    while(t--)    {        __int64 n,nn;        scanf("%I64d",&n);        n++;        int count=0;        nn=sqrt(n);        for(int i=2;i<=nn;i++)            if(n%i==0)count++;        printf("%d\n",count);    }    return 0;}