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  • POJ 2282 数位DP

    链接:

    http://poj.org/problem?id=2282

    题意:

    给你一个区间a,b,问a到b之间每个数字出现了多少次

    题解:

    看过算法设计与分析的人都很熟悉这道题,毕竟是课后练习的第一道,感觉用数位dp比模拟更好理解啊

    dp[pos][sta]表示到从最低位到第pos位,第pos位为sta(0<=sta<=9)时的 num的个数。

    代码:

    31 int a[20];
    32 int dp[20][10];
    33 int num;
    34 
    35 int dfs(int pos, int sta, bool lead, bool limit) {
    36     if (pos == -1) return sta;
    37     if (!lead && !limit && dp[pos][sta] != -1) return dp[pos][sta];
    38     int up = limit ? a[pos] : 9;
    39     int res = 0;
    40     rep(i, 0, up + 1) {
    41         if (lead && i == 0) res += dfs(pos - 1, sta, true, limit && i == a[pos]);
    42         else res += dfs(pos - 1, sta + (i == num), false, limit && i == a[pos]);
    43     }
    44     if (!lead && !limit) dp[pos][sta] = res;
    45     return res;
    46 }
    47 
    48 int solve(int x) {
    49     int pos = 0;
    50     while (x) {
    51         a[pos++] = x % 10;
    52         x /= 10;
    53     }
    54     return dfs(pos - 1, 0, true, true);
    55 }
    56 
    57 int main() {
    58     ios::sync_with_stdio(false), cin.tie(0);
    59     int a, b;
    60     memset(dp, -1, sizeof(dp));
    61     while (cin >> a >> b, a) {
    62         if (a > b) swap(a, b);
    63         num = 0;
    64         cout << solve(b) - solve(a - 1);
    65         for (num = 1; num <= 9; num++)
    66             cout << ' ' << solve(b) - solve(a - 1);
    67         cout << endl;
    68     }
    69     return 0;
    70 }
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  • 原文地址:https://www.cnblogs.com/baocong/p/6811611.html
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