An easy problem+数学推理
2013-04-25 12:05
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An easy problem
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4575 Accepted Submission(s): 1117
[align=left]Problem Description[/align]
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 ?
[align=left]Input[/align]
The first line contain a T(T <= 2000) . followed by T lines ,each line contain an integer N (0<=N <= 1010).
[align=left]Output[/align]
For each case, output the number of ways in one line.
[align=left]Sample Input[/align]
2 1 3
[align=left]Sample Output[/align]
0 1思路:数学推理 [code]#include<stdio.h> #include<math.h> int main() { int n,k; __int64 i,j,m; scanf("%d",&n); while(n--) { scanf("%I64d",&m); m=m+1; j=sqrt(m); k=0; for(i=2;i<=j;i++) { if(m%i==0) k++; } printf("%d\n",k); } return 0; }
[/code]
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