A - Diagonal String

水题：记录直接输出就好

//
//  main.cpp
//  atcoder
//
//  Created by cfhaiteeh on 11/03/2018.
//  Copyright © 2018 cfhaiteeh. All rights reserved.
//
 
#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
char ch[4][4];
int main() {
    for(int i=1;i<=3;i++)scanf("%s",ch[i]+1);
    for(int i=1;i<=3;i++)printf("%c",ch[i][i]);
    printf("\n");
    return 0;
}

B - Palindromic Numbers

数据较小，判断记录总的多少就好

//
//  main.cpp
//  atcoder
//
//  Created by cfhaiteeh on 11/03/2018.
//  Copyright © 2018 cfhaiteeh. All rights reserved.
//

#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
char ch[4][4];
int ai[123456];
  int ci[6];
int main() {
    for(int i=10000;i<=99999;i++){
      
        int st=0;
        int t=i;
        while(t>0){
            int p=t%10;
            t=t/10;
            ci[++st]=p;
        }
        bool flag=1;
        for(int j=1;j<=2;j++){
            if(ci[j]==ci[5-j+1]){
                continue;
            }
            flag=0;
            break;
        }
        if(flag)ai[i]=1;
    }
    for(int i=10000;i<=99999;i++)ai[i]+=ai[i-1];
    int l,r;
    while(scanf("%d%d",&l,&r)!=EOF){
        printf("%d\n",ai[r]-ai[l-1]);
    }
    return 0;
}

C - Flip,Flip, and Flip......

找规律，特判就行

//
//  main.cpp
//  atcoder
//
//  Created by cfhaiteeh on 11/03/2018.
//  Copyright © 2018 cfhaiteeh. All rights reserved.
//
 
#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#define LL long long
using namespace std;
LL n,m;
LL ans;
int main() {
    while(scanf("%lld%lld",&n,&m)!=EOF){
        if(n>m)swap(n,m);
        if(n==1&&m==1)printf("1\n");
        else {
            if(n==1)printf("%lld\n",m-2);
            else {
                LL ans=(n-2)*(m-2);
                printf("%lld\n",ans);
            }
        }
    }
    return 0;
}

D - Remainder Reminder

对于除数进行枚举，易知当除数为i时，1~n的余数为1，2，。。。。0，1，2.。。。0 循环。所以只要直接判断n里有多少组1.2.。。0即可，对于不能整除时进行特判就行。复杂度为O（n）本人对0也进行了特判，主要感觉0怪怪的。。。//

//  main.cpp
//  atcoder
//
//  Created by cfhaiteeh on 11/03/2018.
//  Copyright © 2018 cfhaiteeh. All rights reserved.
//
 
#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#define LL long long
using namespace std;
LL n,m;
LL ans;
int main() {
    while(scanf("%lld%lld",&n,&m)!=EOF){
        ans=0;
        if(m==0){
            ans=n;
            for(int i=2;i<=n;i++){
                LL p=n/i;
                ans+=p;
            }
            m=1;
        }
       
        
            for(LL i=m+1;i<=n;i++){
                LL p=n/i;
                LL q=(n%i)-m+1;
                if(q<0)q=0;
                LL z=i-m;
                ans=ans+z*(p)+q;
                
            }
            printf("%lld\n",ans);
        
    }
    return 0;
}