1001 (Logical Reasoning)
Topic: The two options are A, B and C. Now they choose the option and the points they get tell you whether they lie or not.
Idea: Judging the upper and lower limits is enough.

#include <iostream>  
#include <cstdio>  
#include <cstring>  
using namespace std;  
const int N=80000+5;  
char a[N],b[N];  
int main()
{  
    int T;  
    scanf("%d",&T);  
    while (T--)
    {  
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        int n,x,y;  
        scanf("%d%d%d",&n,&x,&y);  
        scanf("%s\n%s",a,b);   
        int num=0;  
        int i;
        for(i=0;i<n;i++)
        {  
            if(a[i]==b[i])  num++;  
        }  
        int k=0;  
        if(x>=num&&y>=num)
        {  
            int x1=x-num;
            int y1=y-num;  
            int x2=x1+y1;  
            if(x2>n-num)  k=1;  
        }  
        else if(x>=num&&y<num)
        {  
            int x1=x-y;  
            if(x1>n-num)  k=1;      
        }  
        else if(x<num&&y>=num)
        {  
            int y1=y-x;  
            if(y1>n-num)  k=1;  
        }  
        else if(x<num&&y<num)
        k=0;       
        if(!k) printf("Not lying\n");  
        else printf("Lying\n");        
    }       
    return 0;  
}   

1003
Question meaning: Give a sequence a, each time you can choose any number from the array b to indicate that you can choose any number from the number with the subscript 1 to n (where n, each time you make a new number n will grow) and make a number with the subscript j at the end of the sequence a. The value of this number is a J - J. Now ask what is the maximum number of the newly produced.
Thought: Sort each time to get the maximum sum can be

#include <cstdio>  
#include <algorithm>  
#include <iostream>  
#define maxn 250010   
const int mod=1e9+7;  
using namespace std;  
int a[maxn],b[maxn],c[maxn],d[maxn],n;   
void f()
{
    int i,j;
    for (i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        d[i]=a[i]-i;
    }  
    for(i=1;i<=n;i++)
        scanf("%d",&b[i]);
    c[n]=d[n];
    for(i=n-1;i>0;i--)
    c[i]=max(c[i+1],d[i]);
    sort(b+1,b+1+n);
    int x=c[b[1]];
    int ans=x;
    for(i=1;i<=n;i++)
    c[i]=max(c[i],x-(n+1));
    for(i=2;i<=n;i++)
    {
        x=c[b[i]];
        ans+=x;
        ans%=mod;
    }
    printf("%lld\n", ans);
}
int main()
{
    while(~scanf("%d",&n))
    f();
    return 0;
}

1011
Topic: A coordinate of 200 x 200 has n integer points. Ask how many different regular polygons can be formed.
Idea: Just ask for the square (notice the diamond)

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
const int MAXN = 605;
int mp[MAXN][MAXN];
struct Point{
    int x, y;
}p[MAXN];
int cmp(Point A, Point B)
{
    if(A.x==B.x)return  A.y<B.y;
    return A.x<B.x;
}
int main()
{
    int n;
    while(~scanf("%d", &n))
    {
        memset(mp, 0, sizeof(mp));
        for(int i=0; i<n; i++)
        {
            int x,y;
            scanf("%d%d",&x,&y);
            p[i].x=(x+300);
            p[i].y=(y+300);
            mp[p[i].x][p[i].y]=1;
        }
        sort(p,p+n,cmp);
        int sum=0;
        for(int i=0;i<n;i++)
        {
            for(int j=i+1;j<n;j++)
            {
                int dx=p[i].x-p[j].x;
                int dy=p[j].y-p[i].y;
                if(mp[p[i].x+dy][p[i].y+dx]&&mp[p[j].x+dy][p[j].y+dx]) sum++;
            }
        }
        printf("%d\n",sum/2);
    }
    return 0;
}