题解 | #Number Steps#

#include<stdio.h>
int main() {
    int j = 0, i;
    int a[5000][5000];
    //printf("%d",a[1][1]);
    for (int i = 0; i < 5000; i++)
        for (int j = 0; j < 5000; j++) {
            if (i == j) {
                if (i == 0)
                    continue;

                if (i % 2 == 0)
                    a[i][j] = a[i-1][j-1] + 3;
                else
                    a[i][j] = a[i-1][j-1] + 1;
            } else {
                if (i == j - 2) {
                    if(i==0){
                        a[i][j]=2;
                    }
                    else if (i % 2 == 0)
                        a[i][j] = a[i-1][j-1] + 3;

                    else
                        a[i][j] = a[i-1][j-1] + 1;
                }
            }
        }

    int n=0,x,y;
    //scanf("%d",&n);
    //for(int i=0; i<n;i++){
        scanf("%d %d",&y,&x);
        if(x==0&y==0){
            printf("%d\n",0);
        }
        if(x!=0&y!=0&a[x][y]==0){

            printf("%s\n","No Number");
        }
        else {
            printf("%d\n",a[x][y]);
        }
        
    //}
        
    return 0;
}