上过小学二年级的朋友都能推出piqin-ei*di2。这里我们二分查询一下1到上面一坨就可以了。我想我没必要再讲二分算法的思路#includebits/stdc.husingnamespacestd;intk;intmain(){cink;while(k--){longlongn,e,d;cinnde;longlongsn-e*d2;longlongws/2;longlongl1,rw;boolf0;while(lr){longlongmid(rl)/2;longlongtmid*(s-mid);if(tn){coutmid s-midendl;f1;break;}elseif(tn){rmid-1;}else{lmid1;}}if(!f)coutNOendl;}return0;}有点粗糙但我只能这样勉强完成作业