An easy problem HDU 2132
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 <= 10 10).
Output
For each case, output the number of ways in one line.
Sample Input
2
1
3
Sample Output
0
1
思路：可以轉化為n+1=(i+1)*(j+1)；
#include<stdio.h> #include<math.h> int main() { int T,ans; long long i,a,b; scanf("%d",&T); while(T--) { ans=0; scanf("%lld",&a); a=a+1; b=sqrt(a); for(i=2;i<=b;i++) { if(a%i==0) ans++; } printf("%d\n",ans); } return 0; }
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An easy problem HDU 2132
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