Thinking process

I use a two dimension array to save the code of the highest carpet.
code[x][y] represents the (x, y) point's code of the highest carpet. So the solution comes to me. What i need to do is solving the problem on how to judge the code of the highest carpet. when given (x, y), length on x and height on y, I am sure that each points on (x -> x+length, y + height) is updated their code[each_x][each_y]. so let's do it.

#include<stdio.h>

int code[10][10];

int main() {
    for(int i = 1;i < 10; ++ i) 
        for(int j = 1;j < 10; ++ j)
            code[i][j] = -1;
    
    int n;
    scanf("%d", &n);
    
    for(int i = 1; i <=n ; i ++) {
        int a, b, g, k;
        scanf("%d%d%d%d", &a, &b, &g, &k);
        for(int x = a; x <= a + g; x ++) {
            for(int y = b;y <= b + k; ++ y) {
                code[x][y] = i;
            }
        }
    }
    
    int x,y;
    scanf("%d%d", &x, &y);
    printf("%d", code[x][y]);
    
}

although it can pass the samples, it doesn't work when it comes to the big chess. so quit. have another algorithm to solve the problem.
Please read the description about the title again. I found that it concerns about the only one pair of (x, y). So it's enough to maintain (x, y) poins. When given the (a, b) and g,k respectively on x and y, i can judge whether (x, y) is contained in (a, b) to (a + g, b + k). each a,b,k,g arrays are created by 100,010 of one dimension.It perfectly solves the space overflow. To some extend, I thought it works. so let's code!

#include<stdio.h>

const int N = 100010;
int a[N], b[N], k[N], g[N];

int main() {
    
    int n;
    scanf("%d", &n);
    
    for(int i = 1; i <=n ; i ++) {
        scanf("%d%d%d%d", &a[i], &b[i], &g[i], &k[i]);
    }
    int x,y;
    scanf("%d%d", &x, &y);
    
    int code = -1;
    for(int i = 1;i <= n;i ++) {
        int righttop_x = a[i] + g[i], righttop_y = b[i] + k[i];
        if(righttop_x >= x && a[i] <= x && righttop_y >= y && b[i] <= y)
            code = i;
    }
    printf("%d", code);
}

ok it's good. we also can think about the situation when i reserve the order to check the code of the highest carpet. what will happen? If the carpet covered on the (x, y) point is back-end in the array, program runs fast. If not, runs slowly.