网易2017内推笔试编程题合集二

解题1xi1,000,000,00023010n＊30然后进行行与行之间的xor，其中 1^0=1;所以这道题就被转化成了求矩阵的秩,求法如下。//////Createdbyon//Copyright©20160_0Allrights#include<iostream>usingnamespacestd;#include<vector>#include<algorithm>intgetHighBit(intinthighbit=0while(num)num>>=1;}return}intmain()vector<int>intwhile(cin>>n){colors.clear();intresult=0;inttemp;inti=while(i--)cincolors.push_back}//colors进行从小到大的排序sort(colors.begincolors.end());intbigger,smaller;//biggersmallerbigger=n-1;smaller=bigger-1;while(colors.size()>2)//if(getHighBit(colors[bigger])==getHighBit(colors[smallerinttem=colors[bigger]^colors[smallerif(find(colors.begin(),colors.end(),tem)==colors.end()){colors.push_back(tem);sort(colors.begin(),colors.end}}else}//colors.pop_backbigger=smaller;}cout<<result+2<<endl}}--->b,sum+bpi*b，则，b1，且为必要非充分条件。<=#include#includeint4 intcomp(constvoid*a,constvoid return*(int*)a- intdfs(intpos,longlongsum,longlongintif(sum>pi)elseif(bag[i]==1)elsesum-for(;i<n-1&&bag[i]==bag[i+1];++i);// return int}return}x1-x2*x1-x2y1-y2*y1-y2))的算术平方根123classMain4publicstaticvoidmain(String[]args)5Scannerin=new6intcol=7introw=89int[][]grid=newii<row;%4==0||i4==for(intj j<col;if(j4==0||j4grid[i][j]=}}}{jj<col;%4 2||j4grid[i][j]=}}}}intcount=for(intx=x<row;y 0;y<col;if(grid[x][y]==}}}}}11xx111xx1xx11xxx11x 2 456 usingnamespace891113 vector<int> for(inti= for(inti }vector<int>intmin for(inti={min=vx[i]+vy[i]-}}65

种插入办法可以使新串是一个回文串。如果字符串B插入的位置不同就考虑为不一样的办法。Aaba”，Bb”4种把BA""在字母‘b’之后abba"*在第二个字母'a'"abab"不是回文所以满足条件的答案为21234usingnamespace5boolHuiwen(string6{intfor(int if(str1[i]!=str1[length-returnlength=length-}return}{stringint{count=for(int{}}return}4*x38*x+7100,0001,000,000,007整除的位importimportimportimportimport6publicclasspublicstaticvoidmain(String[]args)Scannerin=newintMap<Long,Integer>map=newHashMap<Long,Integer>(); Queue<Long>queue=new map.put((long)x, long if(n==0){System.out.println(map.get(n)-1); if(map.get(n)>=100001)map.put((4*n+3)%1000000007,}map.put((8*n+7)%1000000007,}}}}}"car"<"carriage"<"cats"<"doggies<"koala""car"<"cats"<"koala"<"doggies"< import2 *CreatedbyGengeon2016-08-publicclassMainpublicstaticvoidmain(String[]args)Scannerscanner newwhile(scanner.hasNext())n=String[]words=newfor(inti 0;i<n;i++)words[i]=}}}privatestaticStringvalidate(String[]words)booleanisABC=booleanisLEN=if(isABC&&isLEN)return}if(isABC)return}if(isLEN)if(isLEN)return}return}staticbooleanisLen(String[]words)booleanresult=for(inti=1;i<words.length;i++)if(words[i].length()<=words[i{result=}}return}staticbooleanisAbc(String[]words)booleanresult for(inti=1;i<words.length;i++)if(words[i].compareTo(words[i1])0)result=}return 5612classMain34publicstaticvoidmain(String[]args)5Scannersc=new67Stringword=89if(isAllUpCase(word)&&isConEql(word)&& publicstaticbooleanisAllUpCase(Stringreturnword.matches("[A-}publicstaticbooleanisConEql(Stringreturn}publicstaticbooleanisThrEql(Stringreturn}}F[0]=F[1]=foreachi≥2:F[i]=F[i-1]+F[i-每一步你可以把当前数字XX-1X+1，现在给你一个数NFibonacci数。3usingnamespacestd;456intFibonacci(int7{8if(N==9returnelseif(N==returnreturn-1)+Fibonacci(N-}int{>> for(inti=0;i<100000; if(Fibonacci(i)-N>{num= 36

if(Fibonacci(num)-N<N-Fibonacci(num-cout<<Fibonacci(num)-cout<<N-Fibonacci(num-return{++7=6现在小易给你nn *Sortn**5importimportpublicclassMainprivatestaticintcheck(int[]if(X[0]>1)return

intif(n==1)return{intsum=for(inti 1;i<n;