CF550C Divisibility by Eight

原题链接

• 题意：给一个数字，要求构造一个数字，只通过删除某些位的数，得到的是能整除 (8) 的数字。
• 题解：
• 代码：

    #include <bits/stdc++.h>
    using namespace std;
    typedef long long LL;
    const int N = 2e5 + 10;
    int dp[N][9];
    struct node {
        int pos, data;
        char ans;
    }pre[N][9];
    char a[N];
    int main() {
        cin >> (a  +1);
        int n = strlen(a + 1);
        dp[1][(a[1] - '0')%8] = 1;
        pre[1][(a[1]-  '0')%8] = {0, (a[1] - '0')%8, a[1]};
        for (int i = 2; i <= n; i ++) {
            for (int j = 0; j < 8; j ++) {
                if (dp[i-1][j]) {
                    dp[i][(j*10 + a[i] - '0')%8] = 1;
                    pre[i][(j*10 + a[i] - '0')%8] = {i-1, j,a[i]};
                    dp[i][j] = 1;
                    //cout << "?";
                    pre[i][j] = pre[i-1][j];
                }
            }
            if (!dp[i][(a[i] - '0')%8]) {
                dp[i][(a[i] - '0')%8] = 1;
                pre[i][(a[i] - '0')%8] = {0, (a[i] - '0')%8, a[i] };
            }
        }
        string ans = "";
        for (int i = 1; i <=n ; i ++) {
            if (dp[i][0]) {
                node now =  pre[i][0];
                int u = now.pos;
                while (u) {
                    ans += now.ans;
                    now = pre[now.pos][now.data];
                    u = now.pos;
                }
                ans += now.ans;
                reverse(ans.begin(), ans.end());
                puts("YES");
                cout << ans << endl;return 0;
            }
        }
        puts("NO");
        return 0;
    }