2013 Multi-University Training Contest 3 (g) The Unsolvable Problem
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题目地址: http://acm.hdu.edu.cn/showproblem.php?pid=4627
[a,b]=ab/(a,b) a,b和一定时,越接近乘积越大,在接近的同时,尽量保持公因子最少。 如果n为2*k+1 那么 取a=2k,b=2k+1 最佳。 如果n为偶数=2k,就要进一步分k的奇偶性了。 n=4k 时 显然2k+2k会很小 2k-1,2k+1 就满足了 (2k-1,2k+1)=(2,2k+1)=1 。 如果n=4k+2 ,逐一实验2k+1,2k+1。 2k 2k+2. 2k-1,2k+3 。发现2k-1,2k+3是最佳的,而且他们已经互素了,再增大差距结果不会优于这组。
小细节是,2=4*0+2. 2*0-1=-1了 这个需要特判 , 然后int会超范围,要用long long
代码:
#include<iostream>using namespace std;typedef long long inta;int main(){ inta T; cin>>T; inta a,b,n; while(T--) { cin>>n; if(n%2==1) { cout<<(n*n-1)/4<<endl; } else if(n%4==0) { cout<<(n*n)/4-1<<endl; } else { if(n==2) cout<<1<<endl; else { inta k=n/2-1; cout<<k*k+2*k-3<<endl; } } }}
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