算法之美――浙大计算机研究生复试上机考试-2009年 - 小熊不去实验室 - 博客频道 - CSDN.NET


XXX定律
对于一个数n,如果是偶数,就把n砍掉一半;如果是奇数,把n变成 3*n+ 1后砍掉一半,直到该数变为1为止。
请计算需要经过几步才能将n变到1,具体可见样例。
  1. int main()  
  2. {  
  3.     int n;  
  4.     while (cin>>n&&n!=0)  
  5.     {  
  6.         int cnt=0;  
  7.         while(n!=1)  
  8.         {  
  9.             n=n%2?(n*3+1)>>1:n>>1;  
  10.             cnt++;  
  11.         }  
  12.         printf("%d\n",cnt);  
  13.     }  
  14. }  
ZOJ
读入一个字符串,字符串中包含ZOJ三个字符,个数不一定相等,按ZOJ的顺序输出,当某个字符用完时,剩下的仍然按照ZOJ的顺序输出。
  1. int main()  
  2. {  
  3.     string str;  
  4.     while(cin>>str&&str!="E")  
  5.     {  
  6.         int i;  
  7.         int count[26];  
  8.         memset(count,0,sizeof(int)*26);  
  9.         for (i=0;i<str.length();i++)  
  10.             count[str[i]-'A']++;  
  11.         int z='Z'-'A',o='O'-'A',j='J'-'A';  
  12.         while (count[z]>0||count[o]>0||count[j]>0)  
  13.         {  
  14.             if(count[z]-->0)  
  15.                 printf("Z");  
  16.             if(count[o]-->0)  
  17.                 printf("O");  
  18.             if(count[j]-->0)  
  19.                 printf("J");  
  20.         }  
  21.         printf("\n");  
  22.     }  
继续xxx定律
当n为3时,我们在验证xxx定律的过程中会得到一个序列,3,5,8,4,2,1,将3称为关键数,5,8,4,2称为覆盖数。现在输入n个数字a[i],根据关键数与覆盖数的理论,我们只需要验证其中部分数就可以确定所有数满足xxx定律,输出输入的n个数中的关键数。如果其中有多个关键数的话按照其输入顺序的逆序输出。

找出直系亲属
如果A,B是C的父母亲,则A,B是C的parent,C是A,B的child,如果A,B是C的(外)祖父,祖母,则A,B是C的grandparent,C是A,B的grandchild,如果A,B是C的(外)曾祖父,曾祖母,则A,B是C的great-grandparent,C是A,B的great-grandchild,之后再多一辈,则在关系上加一个great-。

  1. int Dfs(char a,char b,int level)  
  2. {  
  3.     //分别从a和b往下找  
  4.     if(a=='-'||tree[a][0]=='-'&&tree[a][1]=='-')  
  5.         return INF;  
  6.     if(tree[a][0]==b||tree[a][1]==b)  
  7.         return level+1;  
  8.     int left=Dfs(tree[a][0],b,level+1);  
  9.     int right=Dfs(tree[a][1],b,level+1);  
  10.     return min(left,right);  
  11. }  
  12. int main()  
  13. {  
  14.     string str;  
  15.     int n,m,i,j;  
  16.     while (scanf("%d%d",&n,&m)!=EOF&&!(n==0&&m==0))  
  17.     {  
  18.         memset(tree,'-',sizeof(char)*2*('Z'+10));  
  19.         for(i=0;i<n;i++)  
  20.         {  
  21.             cin>>str;  
  22.             tree[str[0]][0]=str[1];  
  23.             tree[str[0]][1]=str[2];  
  24.         }  
  25.         for(i=0;i<m;i++)  
  26.         {  
  27.             cin>>str;  
  28.             int level1=Dfs(str[0],str[1],0);  
  29.             int level2=Dfs(str[1],str[0],0);  
  30.             int level=level1>=INF?-level2:level1;  
  31.             if(abs(level)>=INF)  
  32.             {  
  33.                 printf("-\n");  
  34.                 continue;  
  35.                 }  
  36.             while (level++<-2)  
  37.                 printf("great-");  
  38.             level--;  
  39.             if(level==-1)  
  40.                 printf("parent\n");  
  41.             else if(level==-2)  
  42.                 printf("grandparent\n");  
  43.   
  44.             while (level-->2)  
  45.                 printf("great-");  
  46.             level++;  
  47.             if (level==2)  
  48.                 printf("grandchild\n");  
  49.             else if(level==1)  
  50.                 printf("child\n");  
  51.         }  
  52.     }  
  53. }  
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