基础语法：

字符串相关

字符串函数库：cstring
获取字符串长度：strlen(s)
字符串数组：使用二维数组表示 a[元素数目][最大长度]，定义时赋值或循环赋值scanf("%s", strs[i])
字符串赋值：a=b→strcpy(a,b)；

算法及数据结构回顾：

题解：

1151

二叉树，LCA，vector与Map应用
参考自:https://blog.csdn.net/qq_40597059/article/details/99699193，利用容器特性和遍历序列相对位置判断

#include<iostream>
#include<map>
#include<vector>
using namespace std;
map<int,int>pos;
vector<int>pre,in;
void lca(int inl,int inr,int preroot,int a,int b)
{
    if(inl>inr) return;
    int inroot=pos[pre[preroot]],ina=pos[a],inb=pos[b];
    if(ina<inroot&&inb<inroot) lca(inl,inroot-1,preroot+1,a,b);
    else if(ina>inroot&&inb>inroot) lca(inroot+1,inr,preroot+inroot-inl+1,a,b);
    else if((ina<inroot&&inb>inroot)||(ina>inroot&&inb<inroot)) printf("LCA of %d and %d is %d.\n",a,b,in[inroot]); 
    else if(ina==inroot)    printf("%d is an ancestor of %d.\n",a,b);
    else printf("%d is an ancestor of %d.\n",b,a);
}
int main()
{
    int m,n;
    scanf("%d%d",&m,&n);
    pre.resize(n+1);
    in.resize(n+1);
    for(int i=1;i<=n;i++) 
    {
        scanf("%d",&in[i]);
        pos[in[i]]=i;
    }
    for(int i=1;i<=n;i++) 
    {
        scanf("%d",&pre[i]);
    }
    while(m--)
    {
        int a,b;
        scanf("%d%d",&a,&b);
        if(pos[a]==0&&pos[b]==0)    printf("ERROR: %d and %d are not found.\n",a,b);
        else if(pos[a]==0)  printf("ERROR: %d is not found.\n",a);
        else if(pos[b]==0)  printf("ERROR: %d is not found.\n",b);
        else lca(1,n,1,a,b);
    }
    return 0;
}  

1042

基础题，字符串操作

#include<cstdio>
#include<cstring> 
int main()
{
    char card[55][5]={" ","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"};
    char temp[55][5];
    int order[54];
    int count,i,j;
    scanf("%d",&count);
    for(i=1;i<=54;i++)  scanf("%d",&order[i]);
//  for(i=0;i<54;i++)   printf("%d",order[i]);
    for (j=1;j<=count;j++)
    {
        for(i=1;i<=54;i++)
        {
            strcpy(temp[order[i]],card[i]);
        }
        for(i=1;i<=54;i++) strcpy(card[i],temp[i]);
    }
    for(i=1;i<=54;i++) 
    {
        if(i!=1) printf(" "); 
        printf("%s",card[i]);
    
    }return 0;
}

1005

基础题

#include<cstdio>
#include<cstring>
int main()
{
    char s[100];
    int sum=0,j=0,ans[100];
    scanf("%s",&s);
    for(int i=0;i<strlen(s);i++) sum+=int(s[i])-48;
    if(sum==0) printf("zero"); 
    while(sum)
    {
        ans[j]=sum%10;
        sum/=10;
        j++;
    }
    while(j--)
    {
        if(ans[j]==0) printf("zero");
        else if(ans[j]==1) printf("one");
        else if(ans[j]==2) printf("two");
        else if(ans[j]==3) printf("three");
        else if(ans[j]==4) printf("four");
        else if(ans[j]==5) printf("five");
        else if(ans[j]==6) printf("six");
        else if(ans[j]==7) printf("seven");
        else if(ans[j]==8) printf("eight");
        else if(ans[j]==9) printf("nine");
        if(j!=0)
            printf(" ");
    }
    
    return 0;
} 

1001

基础题，数据小，感觉我的处理有些复杂了

#include<cstdio>
#include<cmath>
int main()
{
    int a,b,ans,t1,t2,t3;
    scanf("%d %d",&a,&b);
    ans=a+b;
    t1=ans%1000;
    ans/=1000;
    t2=ans%1000;
    ans/=1000;
    t3=ans%1000;
    if(t3!=0)   printf("%d,%03d,%03d",t3,abs(t2),abs(t1));
    else if(t2!=0)  printf("%d,%03d",t2,abs(t1));
    else printf("%d",t1);
    return 0;
}