poj 3208 Apocalypse Someday(数位dp)
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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;}
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