2013年江西理工大学C语言程序设计竞赛（初级组）

ACM ICPC WORLD FINAL

解法：排序大家都知道，去重的话，初学者用数组就好了

?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29

#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--;     } }

木

解法：找规律，前面的n行都是在中间输出*，第n+1行全部输出*，接下来的以中间为对称关系，往两边扩展

?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42

#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; }

我们都是江理人

解法：字符串处理（根据题意）

?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

#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]!='\0';t++)         {             if(a[t]=='1')             printf("love jiangli\n");             if(a[t]=='2')             printf("love xingong\n");         }     }     return 0; }

回文素数

解法：数据不大，当然是先判断是不是回文再判断素数，（这里可以把数字一位一位分解倒着相加看是否相等）

 

?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34

#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);     } }

 兽兽扔铅球

解法：数学题，没什么好说的

?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

#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; }

魔兽争霸

解法：应该是计算斜率了(y2-y1)*(x3-x1)==(y3-y1)*(x2-x1)

?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

#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; }