Nine degrees oj& Beihang University 2009 Machine Questions

Topic One, jobdu1166: iterative seeking cube root

http://ac.jobdu.com/problem.php?pid=1166

Title Description:

The approximation iterative equation for cube root is y (n+1) = y (n) *2/3 + x/(3*y (n) *y (n)), where y0=x. The value of the given x after n iterations cube root.

Input:

Enter more than one set of data.
For each set of rows, enter x N.

Output:

Iterate n times after the cubic root, double the precision, preserving six digits after the decimal point.

Sample input:

3000000 28

Sample output:

144.224957

Topic Analysis:

Recursive or recursive, personal feeling recursion is very good, two variables loop forward can be.

AC Code:

<span style= "FONT-SIZE:18PX;" >/**  * @xiaoran  * recursive or recursive */  #include <iostream> #include <cstdio> #include <map># include<cstring> #include <string> #include <algorithm> #include <queue> #include <vector > #include <stack> #include <cstdlib> #include <cctype> #include <cmath> #define LL Long Longusing namespace Std;int Main () {    double y,x,yy;    int n;    while (cin>>x>>n) {        y=x;        for (int i=1;i<=n;i++) {            yy=y*2/3+x/(3*y*y);            y=yy;        }        printf ("%.6lf\n", y);    } return 0;} </span>

Topic Two, jobdu1167: array sorting

http://ac.jobdu.com/problem.php?pid=1167

Title Description:

Enter the value of an array to find the order in which each value is sorted from small to large.

Input:

Enter more than one set of data.
The first number of inputs per group is the length of an array of n (1<=n<=10000), followed by the number of values in the array, separated by a space.

Output:

The values of each input are arranged in order from small to large (there is no space behind the last digit).

Sample input:

4-3 75 12-3

Sample output:

1 3 2 1

Topic Analysis:

sort + Hash map, the STL map is very useful. Sort the subscript for each value of the record Oh.

AC Code:

<span style= "FONT-SIZE:18PX;" > #include <iostream> #include <algorithm> #include <map>using namespace Std;int a[10001],b[10001] , Cnt[10001];int Main () {    int n;    while (Cin>>n) {for        (int i=0;i<n;i++) {            cin>>a[i];            B[i]=a[i];        }        Sort (a,a+n);        Cnt[0]=1;        map<int,int> num;        Num[a[0]]=1;        for (int i=1;i<n;i++) {            if (a[i]==a[i-1]) num[a[i]]=num[a[i-1]];            else num[a[i]]=num[a[i-1]]+1;        }        for (int i=0;i<n-1;i++) {            cout<<num[b[i]]<< "";        }        cout<<num[b[n-1]]<<endl;    }    return 0;} </span>

Topic three, jobdu1168: Search and delete of string

http://ac.jobdu.com/problem.php?pid=1168

Title Description:

Given a short string (without spaces), and given a number of strings, delete the contained short string in these strings.

Input:

Enter only 1 sets of data.
Enter a short string (with no spaces), and then enter several strings until the end of the file.

Output:

Delete the input short string (case insensitive) and remove the whitespace, output.

Sample input:

in#include int main () {printf ("Hi");}

Sample output:

#cludetma () {prtf ("Hi");}

Topic Analysis:

Don't explain, look at the code. If you don't understand it, it is recommended to study library functions in C + +.

AC Code:

<span style= "FONT-SIZE:18PX;" >/** * @xiaoran * * #include <iostream> #include <cstdio> #include <map> #include <cstring># include<string> #include <algorithm> #include <queue> #include <vector> #include <stack> #include <cstdlib> #include <cctype> #include <cmath> #define LL long longusing namespace Std;int main () {string S1,s,s2;getline (Cin,s), for (int i=0;i<s.size (); i++) {if (Isalpha (S[i])) S[i]=tolower (S[i]);}        int k=s.size (); while (Getline (CIN,S1)) {s2=s1;        for (int i=0;i<s1.size (); i++) {if (Isalpha (S1[i])) S1[i]=tolower (S1[i]);        } String::size_type pos=0;        int ok=0;            while ((Pos=s1.find (S,pos))!=string::npos) {//cout<<pos<<endl;            S1.erase (POS,K);            S2.erase (POS,K);        ++pos;        } for (int i=0;i<s2.size (); i++) {if (s2[i]!= ') cout<<s2[i]; } Cout<<endl;} return 0;} </span>

Nine degrees oj& Beihang University 2009 Machine Questions