HDU 3943 二分+數位dp

題意：

題目鏈接：http://acm.hdu.edu.cn/showproblem.php?pid=3943
在區間[P,Q]中找到第k大的數位上還有x個4以及y個7的數。

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思路：

二分+數位dp。

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代碼：

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;

LL p, q, x, y, k;
int a[25];
LL dp[25][25][25];

LL dfs(int pos, int four, int seven, bool limit) {
    if (pos == -1) return (four == x && seven == y);
    if (four > x || seven > y) return 0;
    if (!limit && dp[pos][four][seven] != -1) return dp[pos][four][seven];
    int up = limit ? a[pos] : 9;
    LL res = 0;
    for (int i = 0; i <= up; i++) {
        res += dfs(pos - 1, four + (i == 4), seven + (i == 7), limit && a[pos] == i);
    }
    if (!limit) dp[pos][four][seven] = res;
    return res;
}

LL cal(LL x) {
    int pos = 0;
    while (x) {
        a[pos++] = x % 10;
        x /= 10;
    }
    return dfs(pos - 1, 0, 0, true);
}

void solve() {
    LL res = -1;
    LL l = p + 1, r = q;
    while (l <= r) {
        LL m = (l + r) >> 1;
        if (cal(m) > k) r = m - 1;
        else if (cal(m) == k) {
            res = m;
            r = m - 1;
        }
        else l = m + 1;
    }
    printf("%I64d\n", res);
}

int main() {
    //freopen("in.txt", "r", stdin);
    int T, cs = 0;
    scanf("%d", &T);
    while (T--) {
        int n;
        memset(dp, -1, sizeof(dp));
        scanf("%I64d%I64d%I64d%I64d%d", &p, &q, &x, &y, &n);
        printf("Case #%d:\n", ++cs);
        LL xx = cal(p), yy = cal(q);
        while (n--) {
            scanf("%I64d", &k);
            if (k > yy - xx) {
                puts("Nya!");
                continue;
            }
            k += xx;
            solve();
        }
    }
    return 0;
}