HDU 2601:An easy problem【数学】
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An easy problem
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 8382 Accepted Submission(s): 2044
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 ?
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
if n=i*j+i+j,then n+1=(i+1)*(j+1)
AC-code:
#include<cstdio>#include<cmath>int main(){int T;long long n,m,i,k;scanf("%d",&T);while(T--){scanf("%lld",&n);n++;k=0;m=sqrt(n);for(i=2;i<=m;i++)if(n%i==0)k++;printf("%lld\n",k);}return 0;}
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