zoj 3816 Generalized Palindromic Number(暴力枚举)

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题目链接：zoj 3816 Generalized Palindromic Number

题目大意：给定n，找一个最大的数x，保证x小于n，并且x为palindromic number

解题思路：枚举前i个放于n相同的数，然后去构造后半部分即可。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
typedef unsigned long long ll;

int n = 0, bit[30], num[30], ans[30];

bool judge(int* a, int* b, int n) {
    for (int i = n - 1; i >= 0; i--) {
        if (a[i] != b[i])
            return a[i] > b[i];
    }
    return false;
}

bool cmp (int* a, int* b, int d) {
    for (int i = 1; i <= d; i++) {
        if (a[n-i] != b[n-i])
            return a[n-i] > b[n-i];
    }
    return false;
}

void dfs (int d, int p, bool flag) {
    if (d < p) {
        if (judge(bit, num, n) && judge(num, ans, n))
            memcpy(ans, num, sizeof(num));
        return;
    }

    int end = (flag ? bit[d] : 9);

    for (int i = end; i >= 0; i--) {
        int v = p;

        if (cmp(ans, num, n - d - 1))
            return;

        num[d] = i;
        if (num[d] != num[d+1]) {
            while (v <= d) {
                num[v++] = num[d];
                dfs(d - 1, v, flag && i == end);
            }
        } else
            dfs(d - 1, p, flag && i == end);
    }
}

ll solve () {
    ll a;
    scanf("%llu", &a);

    memset(ans, 0, sizeof(ans));
    memset(bit, 0, sizeof(bit));
    memset(num, 0, sizeof(num));

    n = 0;
    while (a) {
        bit[n++] = a % 10;
        a /= 10;
    }

    num[n] = 0;
    dfs(n-1, 0, 1);

    ll ret = 0;
    for (int i = n - 1; i >= 0; i--)
        ret = ret * 10 + ans[i];
    return ret;
}

int main () {
    int cas;
    scanf("%d", &cas);
    while (cas--) {
        printf("%llu\n", solve());
    }
    return 0;
}

zoj 3816 Generalized Palindromic Number(暴力枚举)

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原文地址：http://blog.csdn.net/keshuai19940722/article/details/39136645