D - An easy problem
Time Limit:3000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
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 ?
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 <= 10
10).
Output
For each case, output the number of ways in one line.
Sample Input
2
1
3
Sample Output
0
此题是一个数学规律题，（其实我一开始也是想暴力的，TLE了两次，WA了一次，发现必须得用数学公式），不难发现(i+1)*(j+1)=m+1，所以题目变成了求m+1的因数的对数个数
#include <stdio.h>
#include <math.h>
int main()
{ int n;__int64 m,i,j; scanf("%d",&n); while(n--) { scanf("%I64d",&m); m++; int count = 0; j = sqrt(m);for(i = 2;i<=j ; i++)// (i+1)*(j+1) = m+1 { if(m % i == 0) count++; } printf("%d\n",count); } return 0;
}