刷题链接：https://www.patest.cn/contests/pat-a-practise
1001
1 #include <iostream> 2 #include <stdio.h> 3 4 using namespace std; 5 int temp[10]; 6 int main() 7 { 8 int a,b,sum,index; 9 while(scanf("%d%d",&a,&b)!=EOF){ 10 sum=a+b; 11 if(sum<0){ 12 printf("-"); 13 sum=-sum; 14 } 15 index=0; 16 17 while(sum>=1000){ 18 temp[index++]=sum%1000; 19 sum=sum/1000; 20 } 21 printf("%d",sum); 22 for(int i = index-1;i >= 0;i--){ 23 if(temp[i]>99) 24 printf(",%d",temp[i]); 25 else if(temp[i]>9) 26 printf(",0%d",temp[i]); 27 else 28 printf(",00%d",temp[i]); 29 } 30 printf(" "); 31 } 32 return 0; 33 }
数据结构初级入门第一题：https://pta.patest.cn/pta/test/17/exam/4/question/260
1 #include <iostream> 2 #include <stdio.h> 3 4 using namespace std; 5 6 int main() 7 { 8 int a,h,ans,i,j,temp;char c; 9 while(scanf("%d %c",&a,&c)!=EOF){ 10 for(h=1;a+1>=2*h*h;h++); 11 if(a==0)h=0,ans=0; 12 else ans=a-2*(h-1)*(h-1)+1; 13 h=(h-2)*2+1; 14 //printf("h is %d and ans is %d ",h,ans); 15 //下面开始输出 16 temp=h; 17 for(i=0;i<h;i++){ 18 if(i<(h+1)/2){ 19 for(j=0;j<i;j++)printf(" "); 20 for(j=0;j<h-i*2;j++)printf("%c",c); 21 printf(" "); 22 }else{ 23 for(j=0;j<h-i-1;j++)printf(" "); 24 for(j=0;j<2*i+2-h;j++)printf("%c",c); 25 printf(" "); 26 } 27 } 28 printf("%d ",ans); 29 } 30 return 0; 31 }
第二题：思路是素数打表法，请注意做表的时候请将空间开到足够大，否则会出现越界错误。同时相对ACM来说PAT的OJ是有bug的，这也就是说明不管题目你是否会写，请得分，很简单的，至少都有分数的，加油。
#include <iostream> #include <stdio.h> using namespace std; #define NMAX 100000 int num[NMAX]; int main() { int i,j; //思路：采用素数打表法对素数进行处理 for(i=0;i < NMAX-1;i++){ num[i]=0; } for(j=2;j < NMAX-1;j++){ if(num[j]==0)for(int k =2*j;k<NMAX-1;k=k+j)num[k]=1; } //上边首先是对表进行初始化 int n,count; while(scanf("%d",&n)!=EOF){ count=0; for(i=2;i<n-1;i++){ if(num[i]==0&&num[i+2]==0)count++; } printf("%d ",count); } return 0; }
第三题：佷常见的一个题目，但是要考虑到特殊的数据集的时候，你会发现你以前不严谨的代码错误百出。还好是PAT对于结果会给你一定的提升，请考虑提高数据集的复杂性，将问题考虑全面
#include <iostream> #include <stdio.h> using namespace std; int num[103]; void reserve(int *,int,int); int main() { int i,n,offset; while(scanf("%d%d",&n,&offset)!=EOF){ if(n==0)continue; offset%=n; for(i=0;i<n;i++)scanf("%d",&num[i]); reserve(num,0,n-offset-1); reserve(num,n-offset,n-1); reserve(num,0,n-1); for(i=0;i<n;i++) if(i!=n-1) printf("%d ",num[i]); else printf("%d",num[i]); printf(" "); } return 0; } void reserve(int *a,int b,int c){ int temp; if(b>=c)return; for(int i=b;i <= b+(c-b)/2;i++){ temp=a[i]; a[i]=a[c-(i-b)]; a[c-(i-b)]=temp; } }
第四题：这个题真的是靠着PAT的提示给整出来的，这个编译器的是好到爆，希望考试要有这样一个效果就好了
这个题目要特别注意！！！进位问题
#include <iostream> #include <stdio.h> #include <string.h> #include <string> using namespace std; #define NMAX 100 char num1[NMAX],num2[NMAX]; int demo1[10],demo2[10]; int main() { int i,j,add,count;bool flag; while(scanf("%s",num1)!=EOF){ flag=true; for(i=0;i<10;i++)demo1[i]=demo2[i]=0; for(i=0;i<strlen(num1);i++)demo1[num1[i]-'0']++; add=count=0; for(i=strlen(num1)-1;i>=0;i--){ j=num1[i]-'0'; j*=2; num2[count++]=(j+add)%10+'0'; add=(j+add)/10; } if(add!=0)num2[count++]=add%10+'0'; num2[count]=' '; for(i=0;i<strlen(num2);i++)demo2[num2[i]-'0']++; for(i=0;i<10;i++){ if(demo1[i]!=demo2[i]){ flag=false; break; } } if(flag)printf("Yes "); else printf("No "); for(i=count-1;i>=0;i--)printf("%c",num2[i]); printf(" "); } return 0; }
第五题：终于是通关了，这个题目用的最笨的一种方法搞定的。同时也是因为自己没有理解题意，数据量太大，没法调试，卡壳了好久。参考了网上的代码后修改自己的把题目弄出来了。也是，代码这种东西满分不易，且行且珍惜。
#include <iostream> #include <stdio.h> using namespace std; #define MAX 60 #define COUNT 54 int place[MAX]; int temp_num[MAX]; int place_num[MAX]; int main() { int n,temp; char* ans[]={ "TEMP","S1","S2","S3","S4","S5","S6","S7","S8","S9","S10","S11","S12","S13", "H1","H2","H3","H4","H5","H6","H7","H8","H9","H10","H11","H12","H13", "C1","C2","C3","C4","C5","C6","C7","C8","C9","C10","C11","C12","C13", "D1","D2","D3","D4","D5","D6","D7","D8","D9","D10","D11","D12","D13", "J1","J2"}; while(scanf("%d",&n)!=EOF){ for(int i =1;i <= COUNT;i++){ scanf("%d",&temp_num[i]); place[i]=place_num[i]=i; } for(int j=1;j <= n;j++){ for(int t=1;t<=COUNT;t++){ place_num[temp_num[t]]=place[t]; } for(int t=1;t<=COUNT;t++){ place[t]=place_num[t]; } } //最后是输出结果 for(int i = 1;i<=COUNT;i++) if(i!=COUNT)printf("%s ",ans[place_num[i]]); else printf("%s",ans[place_num[i]]); printf(" "); } return 0; }
参考代码：
1 参考代码 2 #include <cstdio> 3 #include <cstring> 4 #include <string> 5 #include <queue> 6 #include <cmath> 7 #include <iostream> 8 using namespace std; 9 char f[5]={'S','H','C','D','J'}; 10 int main(){ 11 //freopen("D:\INPUT.txt","r",stdin); 12 int time; 13 scanf("%d",&time); 14 int card1[55],card2[55],i,j; 15 for(i=0;i<=54;i++){ 16 card1[i]=i; 17 } 18 int order[55]; 19 for(i=1;i<=54;i++){ 20 scanf("%d",&order[i]); 21 } 22 for(i=1;i<=time;i++){ 23 for(j=1;j<=54;j++){ 24 card2[order[j]]=card1[j]; 25 } 26 for(j=1;j<=54;j++){ 27 card1[j]=card2[j]; 28 } 29 } 30 printf("%c%d",f[(card1[1]-1)/13],card1[1]%13==0?13:card1[1]%13); 31 for(i=2;i<=54;i++){ 32 printf(" %c%d",f[(card1[i]-1)/13],card1[i]%13==0?13:card1[i]%13); 33 } 34 printf(" "); 35 return 0; 36 }
牛客网二叉搜索树：https://www.nowcoder.com/practice/3d6dd9a58d5246f29f71683346bb8f1b?tpId=40&tqId=21455&tPage=7&rp=7&ru=/ta/kaoyan&qru=/ta/kaoyan/question-ranking
参考博客：http://www.cnblogs.com/rickhsg/p/3650875.html
自己的题解：
1 #include <stdio.h> 2 #include <string> 3 #include <string.h> 4 #include <stdlib.h> 5 6 typedef struct Node{ 7 int key; 8 struct Node *left; 9 struct Node *right; 10 }Node; 11 ///申请一个树的根节点，并将其初始设置为空 12 Node * Tree=NULL,* p,* temp_p; 13 char tempa[12],tempb[12],temp_str[12]; 14 15 int length,i; 16 ///先将对应的二叉树构造出来 17 void Creat(char* temp){ 18 for(i=0;i<strlen(temp);i++){ 19 p=(Node *)malloc(sizeof(Node)); 20 p->key=temp[i]-'0'; 21 p->left=p->right=NULL; 22 23 if(Tree==NULL){ 24 Tree=p; 25 }else{ 26 temp_p=Tree; 27 while(1){ 28 if(temp_p->key==p->key)break; 29 else if(temp_p->key<p->key){ 30 if(temp_p->right==NULL){ 31 temp_p->right=p; 32 break; 33 }else 34 temp_p=temp_p->right; 35 }else 36 if(temp_p->key>p->key){ 37 if(temp_p->left==NULL){ 38 temp_p->left=p; 39 break; 40 }else 41 temp_p=temp_p->left; 42 } 43 } 44 } 45 } 46 } 47 48 void pre_order(Node* Tree){ 49 ///使用前应当将长度设置为0初始化 50 if(Tree==NULL)return; 51 else{ 52 temp_str[length++]=(char)(Tree->key+'0'); 53 //printf("+%c",temp_str[length-1]); 54 pre_order(Tree->left); 55 pre_order(Tree->right); 56 } 57 } 58 int main() 59 { 60 int T; 61 while(scanf("%d",&T)!=EOF){ 62 if(0==T)break; 63 scanf("%s",tempa); 64 Tree=NULL; 65 ///先将对应的二叉树构造出来 66 Creat(tempa); 67 68 ///对构造好的排序二叉树进行前序遍历进行唯一性标识 69 length=0; 70 pre_order(Tree); 71 strcpy(tempa,temp_str); 72 73 ///上边已经将二叉树构造好了，现在只要进行一个判断的工作就ok 74 while(T--){ 75 scanf("%s",tempb); 76 Tree=NULL; 77 78 ///先将对应的二叉树构造出来 79 Creat(tempb); 80 81 ///对构造好的排序二叉树进行前序遍历进行唯一性标识 82 length=0; 83 pre_order(Tree); 84 strcpy(tempb,temp_str); 85 86 if(strcmp(tempa,tempb))printf("NO "); 87 else printf("YES "); 88 } 89 } 90 return 0; 91 }
题解思路：https://www.nowcoder.com/questionTerminal/3d6dd9a58d5246f29f71683346bb8f1b