A - An easy problem
Time Limit:3000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
Submit
Status
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 1
考察思维，两层for循环会超时；
i*j+i+j两边同时加1得到（i+1）*(j+1)=n+1;转化为求n+1的因子个数；
一个for循环从2开始到sqrt(n+1）来判断能否被n+1除尽；
要用long long 否则会错；
代码：
#include<stdio.h>
#include<math.h>
int main()
{int t;long long n;scanf("%d",&t);while(t--){long long cnt=0;scanf("%lld",&n);int m=sqrt(n+1);for(int i=2;i<=m;i++){if((n+1)%i==0)cnt++;}printf("%lld\n",cnt);}return 0;}