UVa 10673 - Play with Floor and Ceil
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题目:方程 x = p * floor(x÷k)+ q * ceil(x÷k)给定x和k,求一组成立的p与q。
分析:数论。设 x = m*k + n(其中,n = x mod k);
若n = 0:x = m*k,floor(x÷k)= ceil(x÷k)= m,x = p * m + q * m,令p = k,q = 0即可;
若n ≠ 0:x = m*k + n,ceil(x÷k)= m+1,floor(x÷k) = m,x = (p+q)* m + q,
此时,令p+q = k,q = n即可(因为n < k,所以一定成立),得p = k - n即可。
说明:当x整除k时,p和q可以互换。
#include <iostream>#include <cstdlib>#include <cstdio>using namespace std;int main(){int n,x,k;while (~scanf("%d",&n))for (int t = 0 ; t < n ; ++ t) {scanf("%d%d",&x,&k);if (x%k)printf("%d %d\n",k-x%k,x%k);else printf("0 %d\n",k);}return 0;}
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