摘要： Problem D Time Limit : 2000/1000ms (Java/Other) Memory Limit : 131072/65536K (Java/Other)Total Submission(s) : 11 Accepted Submission(s) : 4Problem Description Little Valentine liked playing with binary trees very much. Her favorite game was constructing randomly looking binary trees with capita... 阅读全文

摘要： Online Judge Problem Set Authors Online Contests User Web BoardHome PageF.A.QsStatistical Charts ProblemsSubmit ProblemOnline StatusProb.ID: RegisterU... 阅读全文

摘要： Red and BlackTime Limit: 1000MS Memory Limit: 30000K Total Submissions: 17584 Accepted: 9279 DescriptionThere is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a tile, he can move to one of four adjacent tiles. But h.. 阅读全文

摘要： /*猜想对于任意大于1的自然数n，若n为奇数，则将n变为3n+1，否则将n变为一半，求变换的次数 其中n<10^9开始不用长整形的时候会溢出，但是还可以用double解决这个问题刘汝佳教我的*//*//法1 长整形法#include<iostream>#include<math.h>using namespace std;int main(){ __int64 n,count = 0; while(scanf("%I64d",&n)!=EOF) { while(n>1) { if(n%2==1) ... 阅读全文

摘要： #include<iostream>#include<stdio.h>#include<malloc.h>using namespace std;#define MaxSize 100#define MaxWidth 40typedef char ElemType;typedef struct Node2{ char data;//数据域 struct Node2*lchild,*rchild;//左指针域，右指针域}BTNode;void CreateBTNode(BTNode *&b,char *str)//由str串创建二叉链{ BTNode 阅读全文

摘要： /*最长回文子串*/#include<iostream>#include<string.h>#include<ctype.h>using namespace std;#define Max 5000 +10char buf[Max],s[Max];int p[Max];int main(){ int n,m=0,max=0; int i,j; //int k; int x,y; fgets(buf,sizeof(s),stdin); n=strlen(buf); for(i=0;i<n;i++) if(isalpha(buf[i])) ... 阅读全文

摘要： #include<iostream>using namespace std;#define maxvex 100typedef struct{ int a[maxvex][maxvex]; int vex;}Graph;int visiteded[maxvex];void DFS(Graph G,int v){ printf("v%d ",v+1); visiteded[v]=1; for(int j=0;j<G.vex;j++) if(G.a[v][j]==1&&visiteded[j]==0) DFS(G,j);}void ... 阅读全文

摘要： /*不用数组也行,scanf返回成功输入变量的个数，按回车+ctrl+z+回车*/#include<iostream>using namespace std;int main(){ int x,n=0,min,max,s=0; scanf("%d",&x); s=min=max=x; n=1; while(scanf("%d",&x)==1) { s+=x; if(x<min)min=x; if(x>max)max=x; n++; } printf("%d %d %.3lf\n",min,m.. 阅读全文

摘要： /*交换两个数不用中间变量*/#include<iostream>using namespace std;int main(){ int a,b; scanf("%d%d",&a,&b); a=a+b; b=a-b; a=a-b; printf("%d %d\n",a,b); return 0;} 阅读全文

摘要： Leftmost DigitTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 8784 Accepted Submission(s): 3391Problem DescriptionGiven a positive integer N, you should output the leftmost digit of N^N. InputThe input contains several test cases. The first ... 阅读全文