HDU 3943 K-th Nya Number

Problem Description
Arcueid likes nya number very much.
A nya number is the number which has exactly X fours and Y sevens(If X=2 and Y=3 , 172441277 and 47770142 are nya numbers.But 14777 is not a nya number ,because it has only 1 four).
Now, Arcueid wants to know the K-th nya number which is greater than P and not greater than Q.

Input
The first line contains a positive integer T (T<=100), indicates there are T test cases.
The second line contains 4 non-negative integers: P,Q,X and Y separated by spaces.
( 0<=X+Y<=20 , 0< P<=Q <2^63)
The third line contains an integer N(1<=N<=100).
Then here comes N queries.
Each of them contains an integer K_i (0

#include <cmath>#include <ctime>#include <cstdio>#include <cstdlib>#include <cstring>#include <iostream>#include <string>#include <vector>#include <deque>#include <list>#include <queue>#include <stack>#include <map>#include <set>#include <numeric>#include <utility>#include <algorithm>#include <functional>using namespace std;typedef long long ll;const int maxn = 30;ll dp[maxn][10][22][22];int T, X, Y, N;ll P, Q, K;void init() {    memset(dp, 0, sizeof(dp));    for(int i = 0; i <= 9; ++i) {        if(i == 4) {            dp[1][4][1][0] = 1;        } else if(i == 7) {            dp[1][7][0][1] = 1;        } else {            dp[1][i][0][0] = 1;        }    }    for(int i = 2; i <= 20; ++i) {        for(int j = 0; j <= 9; ++j) {            for(int x1 = 0; x1 <= 20; ++x1) {                for(int x2 = 0; x2 <= 20; ++x2) {                    for(int k = 0; k <= 9; ++k) {                        if(j == 4) {                            if(x1 > 0) {                                dp[i][j][x1][x2] += dp[i-1][k][x1-1][x2];                            } else {                                dp[i][j][x1][x2] = 0;                            }                        } else if(j == 7) {                            if(x2 > 0) {                                dp[i][j][x1][x2] += dp[i-1][k][x1][x2-1];                            } else {                                dp[i][j][x1][x2] = 0;                            }                        } else {                            dp[i][j][x1][x2] += dp[i-1][k][x1][x2];                        }                    }                }            }        }    }    return ;}int bit[30], blen;bool check(int x, int y) {    for(int i = 0; i < blen; ++i) {        if(bit[i] == 4) x--;        if(bit[i] == 7) y--;    }    return x == 0 && y == 0;}ll dfs(ll pos, int num4, int num7) {    if(pos < 0) {        return check(X, Y) ? 1 : 0;    }    if(num4 < 0) return 0;    if(num7 < 0) return 0;    ll ret = 0;    for(int i = 0; i < bit[pos]; ++i) {        ret += dp[pos+1][i][num4][num7];    }    if(bit[pos] == 4) num4--;    if(bit[pos] == 7) num7--;    ret += dfs(pos - 1, num4, num7);    return ret;}ll solve(ll x) {    blen = 0;    while(x) {        bit[blen++] = x % 10;        x /= 10;    }    return dfs(blen-1, X, Y);}int main() {    //freopen("aa.in", "r", stdin);    init();    int kcase = 0;    ll n1, n2, ans;    scanf("%d", &T);    while(T--) {        kcase++;        scanf("%I64d %I64d %d %d", &P, &Q, &X, &Y);        n1 = solve(P);        scanf("%d", &N);        printf("Case #%d:\n", kcase);        for(int i = 1; i <= N; ++i) {            scanf("%I64d\n", &K);            ll l = P + 1, r = Q, mid;            ans = -1;            while(l <= r) {                mid = (l + r) / 2;                n2 = solve(mid);                if(n2 - n1 >= K) {                    if(ans == -1) ans = mid;                    else ans = min(ans, mid);                    r = mid - 1;                } else {                    l = mid + 1;                }            }            if(ans == -1) {                printf("Nya!\n");            } else {                printf("%I64d\n", ans);            }        }    }    return 0;}