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poj 3208 Apocalypse Someday(数位dp)

题目链接:poj 3208 Apocalypse Someday

题目大意:给定n,输出第n大包含666的数字。

解题思路:数位dp,用类似AC自动机的思想进行转移。首先dp[i][j]表示说i位最后有j个连续6的情况数,这个预处理出

来。那么dp[i][3]即为i位有多少个满足的数。给定n,先确定位数d。然后从最高位向下判断,一开始肯定是需要3个连续

的6,所以u为3,然后根据后面添加的数字动态修改u值进行判断。

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

using namespace std;
typedef long long ll;
const int maxd = 10;
const int status = 4;
const int g[4][12] = {
    {0, 0, 0, 0, 0, 0, 0, 0, 0, 0},
    {3, 3, 3, 3, 3, 3, 0, 3, 3, 3},
    {3, 3, 3, 3, 3, 3, 1, 3, 3, 3},
    {3, 3, 3, 3, 3, 3, 2, 3, 3, 3}};

ll n, dp[maxd+5][status];

int main () {
    memset(dp, 0, sizeof(dp));
    dp[0][0] = 1;
    for (int i = 1; i <= maxd; i++) {
        dp[i][0] = (dp[i-1][0] + dp[i-1][1] + dp[i-1][2]) * 9;
        dp[i][1] = dp[i-1][0];
        dp[i][2] = dp[i-1][1];
        dp[i][3] = dp[i-1][3] * 10 + dp[i-1][2];
    }

    int cas;
    scanf("%d", &cas);
    while (cas--) {
        scanf("%lld", &n);

        int d = 0, u = 3;
        while (dp[d][3] < n)
            d++;

        while (d) {

            ll k = 0;

//            printf("%d:\n", d);
            for (int i = 0; i < 10; i++) {
                ll tmp = 0;

                for (int j = 3; j >= g[u][i]; j--)
                    tmp += dp[d-1][j];

//                printf("n:%lld status: %d %d %lld\n", n, u, i, tmp);
                if (k + tmp >= n) {
                    printf("%d", i);
                    u = g[u][i];
                    break;
                }
                k += tmp;
            }
            n -= k;
            d--;
        }
        printf("\n");
    }
    return 0;
}

poj 3208 Apocalypse Someday(数位dp)