题意: 给出n, k,求 分析: 假设,则k mod (i+1) = k - (i+1)*p = k - i*p - p = k mod i - p 则对于某个区间,i∈[l, r],k/i的整数部分p相同,则其余数成等差数列,公差为-p 然后我想到了做莫比乌斯反演时候有个分块加速,在区间[i, n / (n / i)],n/i的整数部分相同,于是有了这份代码. #include <cstdio> #include <algorithm> using namespace std;…
UVa10025 ? 1 ? 2 ? ... ? n = k problem The problem Given the following formula, one can set operators '+' or '-' instead of each '?', in order to obtain a given k? 1 ? 2 ? ... ? n = k For example: to obtain k = 12 , the expression to be used will be:…