HDU 2601 An easy problem 因式分解


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..
HDU 2601 An easy problem 因式分解

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
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 10).
Output FZ??">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i0,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

分类:默认分类 时间:2015-02-27 人气:8


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