2013 C Language Program design competition (junior group), Jiangxi University of Technology

ACM ICPC World FINAL

Solution: sort Everybody knows, go to heavy words, beginners with array is good

#include <algorithm> #include <iostream>using namespace std;int main () {    int a,b,c[100],i,d[31];    cin>>a;    While (a>0)    {        cin>>b;        For (i=0;i<31;i++)            d[i]=0;        For (i=0;i<b;i++)        {            cin>>c[i];        }         For (i=0;i<b;i++)         {             d[c[i]]++;         }         For (i=0;i<31;i++)         {             if (d[i]!=0)                cout<<i<< "";         }         cout<<endl;         a--;}    }

Wood

Solution: to find the law, the front n lines are in the middle output *, the n+1 line all output *, the next with the middle as a symmetric relationship, to expand on both sides

 #include <bits/stdc++.h>using namespace Std;int main () {int n;            While (cin>>n&&n) {for (int. i=1;i<=2*n+1;i++) {for (int j=1;j<=2*n+1;j++)                {if (j==n+1) {cout<< "*";                } else if (i==n+1) {cout<< "*";                    } else if (i>n+1) {int pos=i-(n+1);                    cout<<pos<<endl; If (n+1-pos==j| |                    N+1+pos==j) {cout<< "*";                } else {cout<< ".";}            } else {cout<< ".";}}        cout<<endl; }} return 0;} 

We're All Jiang.

Solution: string processing (according to test INSTRUCTIONS)

#include <stdio.h>int main () {    int n,t,i;    Char a[1000];    scanf ("%d", &n);    GetChar ();    While (n--)    {        i=0;              Gets (a);        For (t=0;a[t]!= ' + '; t++)        {            if (a[t]== ' 1 ')             printf ("love jiangli\n");            If (a[t]== ' 2 ')            printf ("love xingong\n");        }    }    Return 0;}

Palindrome Prime

Solution: the data is not small, of course, the first judge is not a palindrome and then judge the prime, (here can be a number a decomposition upside down to see if it is Equal)

#include <stdio.h> int main () {     int m,n,c,b,k,p,q,r;     While (scanf ("%d%d", &m,&n)!=eof)     {         if (m==0&&n==0) break             ;         r=0;         For (k=m; k>=m&&k<=n; k++)         {             b=0;             p=k;             While (k>0)             {                 c=k%10;                 b=b*10+c;                 k=k/10;             }             If (b==p)             {for                 (q=2; q<p; q++)                     if (p%q==0) break                         ;                 If (q==p)                 {                     r=r+1;                 }             }             k=p;         }         printf ("%d\n", r);     } }  

Beast and Beast throw shot

Solution: Math problem, Nothing to say

#include <stdio.h> #include <math.h>int main () {    int n;    Float h,a,l;    While (scanf ("%d", &n)!=eof)    {while        (n--)        {            scanf ("%f%f", &h,&a);            L=h/tan (a);            printf ("%.3f\n", l);        }    }    Return 0;}

Warcraft

Solution: it should be calculated slope (y2-y1) * (x3-x1) = = (y3-y1) * (x2-x1)

#include <stdio.h>int main () {    double x1,x2,x3,y1,y2,y3;    int n;    While (scanf ("%d", &n)!=eof)    {for        (int i=0;i<n;i++)        {            scanf ("%lf%lf%lf%lf%lf%lf",& x1,&y1,&x2,&y2,&x3,&y3);            If ((y2-y1) * (x3-x1) = = (y3-y1) * (x2-x1) && (x2-x1) * (x2-x1) >= (x3-x1) * (x3-x1) && (y2-y1) * (y2-y1) >= (y3-y1) * (y3-y1))            printf ("yes\n");            else printf ("no\n");        }    }    Return 0;}

　　

2013 C Language Program design competition (junior group), Jiangxi University of Technology