An easy problem
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5726 Accepted Submission(s): 1394
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.
Input The first line contain a T(T <= 2000) . followed by T lines ,each line contain an integer N (0<=N <= 10 10).
Output FZ http://www.2cto.com/kf/ware/vc/" target="_blank" class="keylink">vciBlYWNoIGNhc2UsIG91dHB1dCB0aGUgbnVtYmVyIG9mIHdheXMgaW4gb25lIGxpbmUuCgogCjxicj4KClNhbXBsZSBJbnB1dAoKPHByZSBjbGFzcz0="brush:java;">2 1 3
Sample Output
0 1
Author Teddy
Source HDU 1st “Vegetable-Birds Cup” Programming Open Contest
i*j+i+j=n => (i+1)*(j+1)=n+1 所以只要求出(n+1)/(i+1)求出有多少j符合情况就行了。题目中要求j是大于0的，所以要判断求出都j是否大于0，可是一旦加上这个判断，就会超时。 通过证明，这个判断确实是多余的，因为一旦(n+1)%(i+1)==0，就能求出j+1,因为i>0，所以for循环是从2开始的（因为是（i+1）），而i<=j，所以只需要判断到根号n就可以了，因为到了根号n以后情况就重复了，如2*6与6*2只有一种符合情况。那么如果j只有两种情况了，一种是j==0，一种是j>0，显然我们求的是j>0都情况，那么j=0的时候，也就是(i+1）*（0+1）=n+1，可以求出i==n，而for循环只循环到根号n，所以这种情况不可能发生。说的有点罗嗦了。 不过即使是这样，也跑了2000多ms。另外是不是只要求出n+1所有的因子小于等于根号n+1的个数就是次题都解？
//2390MS 228K #include#include int main() { int t; scanf("%d",&t); while(t--) { __int64 n,nn; scanf("%I64d",&n); n++; int count=0; nn=sqrt(n); for(int i=2;i<=nn;i++) if(n%i==0)count++; printf("%d\n",count); } return 0; }