Codeforces Round #698 (Div. 2) B. Nezzar and Lucky Number

题面

暴力打表

根据某位大佬的反馈, 本题完全暴力也是能过的, 但是我就是想打表. 当然, 只需要打一部分的表就可以了, 我们可以证明当a_i >= 10 * d时a_i必然可以分解成lucky number的和. 所以我们只需要打到 10 * d 就可以了.

结论证明

我们设a_i = 10 * d + k, 那么显然我们只需要考虑k能否分解了, 令k = x * 10 + y. 当10 * d <= a_i < 11 * d时, 十位上就是d, 显然成立. 当a_i >= 11 * d时, 我们可以将a_i分解成10 * d + y和x * 10 + d, 显然成立.

完整AC代码

这里打表打到了11 * b(~~因为是凭感觉写的, 当时并没有严谨的证明~~)

点我查看完整代码

#include<cstdio>
#include<cstring>
int a[10][100] = {{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,1,0,0,1,0,0,1,0,0,1,1,0,1,1,0,1,1,0,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,1,0,0,0,1,0,0,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,1,0,0,0,0,1,0,0,0,0,1,0,0,0,0,1,0,0,0,0,1,0,0,0,0,1,0,0,0,0,1,0,0,0,0,1,0,0,0,0,1,0,0,0,0,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,1,0,1,0,0,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,1,0,0,0,0,0,0,1,0,0,1,0,0,0,1,0,0,1,0,0,1,1,0,0,1,0,0,1,1,0,1,1,0,0,1,1,0,1,1,0,1,1,1,0,1,1,0,1,1,1,1,1,1,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,1,0,1,0,0,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,1,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,1,1,0,0,0,0,0,0,0,1,1,1,0,0,0,0,0,0,1,1,1,1,0,0,0,0,0,1,1,1,1,1,0,0,0,0,1,1,1,1,1,1,0,0,0,1,1,1,1,1,1,1,0,0,1,1,1,1,1,1,1,1,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0}};
int main(){
    int t;
    scanf("%d", &t);
    while(t--){
        int x, n;
        scanf("%d%d", &x, &n);
        while(x--){
            int b;
            scanf("%d", &b);
            if (b >= 11 * n || a[n][b])	printf("YES\n");
            else	printf("NO\n");
        }
    }
    return 0;
}