NYOJ216A problem is easy
A problem is easy
时间限制:1000 ms | 内存限制:65535 KB
难度:3
- 描述
- 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 ?
- 输入
- The first line contain a T(T <= 2000) . followed by T lines ,each line contain an integer N (0<=N <= 10^11).
- 输出
- For each case, output the number of ways in one line
- 样例输入
-
2 1 3
- 样例输出
-
0 1
#include<stdio.h> #include<math.h> int main() { int n,cnt,i,j,x; scanf("%d",&n); while(n--) { cnt=0; scanf("%d",&x); for(i=2;i<=sqrt(x+1);i++) if((x+1)%i==0) cnt++; printf("%d\n",cnt); } return 0; }
posted on 2012-08-02 14:36 LinuxPanda 阅读(172) 评论(0) 编辑 收藏 举报