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
分析：
i*j+i+j=n<=> (i+1)*(j+1)=n+1;
因此只要从2到sqrt(n+1)找就可以了。
如果(n+1) % i == 0,那么就说明找到一个。
查看代码---运行号：252167----结果：Accepted
运行时间：
2012-10-05 11:22:30 | 运行人:
huangyibiao
01.
#include <iostream>
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#include <cmath>
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using namespace std;
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int main()
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{
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int testNum;
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cin >> testNum;
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while (testNum--)
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{
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int count = 0;
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long n;
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cin >> n;
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long s = (long)(sqrt(n+1));
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for (long i = 2; i <= s; i++)
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{
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if ((n+1) % i == 0)
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{
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count ++;
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}
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}
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cout << count << endl;
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}
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return 0;
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}
