HDU 3943 (二分+数位DP)

K-th Nya Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 125536/65536 K (Java/Others)
Total Submission(s): 2762    Accepted Submission(s): 869

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<K_i <2^63).

 

Output

For each test case, display its case number and then print N lines.
For each query, output a line contains an integer number, representing the K_i-th nya number in (P,Q].
If there is no such number,please output "Nya!"(without the quotes).

 

Sample Input

138 400 1 11012345678910

 

Sample Output

Case #1:4774147174247274347374Nya!Nya!

 

题意:n个询问,每次求出区间中第k大的有x个4,y个7的数.

先数位DP预处理某一个长度a个4,b个7的数字的个数,然后对于每一个询问二分查找

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#include <cmath>using namespace std;long long dp[22][22][22];long long l, r;int x, y;long long n;int cnt, bit[22];long long f (long long num) {    cnt = 0;    while (num) {        bit[++cnt] = num%10;        num /= 10;    }    long long m1 = x, m2 = y;    long long ans = 0;    for (int i = cnt; i >= 1; i--) {        for (int j = 0; j < bit[i]; j++) {            if (j == 4) {                if (m1)                    ans += dp[i-1][m1-1][m2];            }            else if (j == 7) {                if (m2)                    ans += dp[i-1][m1][m2-1];            }            else                ans += dp[i-1][m1][m2];        }        if (bit[i] == 4)            m1--;        else if (bit[i] == 7)            m2--;        if (m1 < 0 || m2 < 0)            return ans;    }    if (m1 == 0 && m2 == 0)        ans++;    return ans;}void init () {    dp[0][0][0] = 1;    for (int i = 1; i < 22; i++) {        for (int m1 = 0; m1 <= 20; m1++) {            for (int m2 = 0; m2 <= 20; m2++) {                dp[i][m1][m2] += 8*dp[i-1][m1][m2];//01235689                if (m1)                    dp[i][m1][m2] += dp[i-1][m1-1][m2];//4                if (m2)                    dp[i][m1][m2] += dp[i-1][m1][m2-1];//7            }        }    }}int main () {    //freopen ("in.txt", "r", stdin);    int t, kase = 0;    memset (dp, 0, sizeof dp);    init ();    scanf ("%d", &t);    while (t--) {        printf ("Case #%d:\n", ++kase);        scanf ("%lld%lld%d%d", &l, &r, &x, &y);        long long pre = f (l); //cout << pre << endl;        int q;        scanf ("%d", &q);        while (q--) {            scanf ("%lld", &n);            long long ll = l, rr = r;            while (rr-ll > 1) {                long long mid = (ll+rr)>>1;                long long cur = f (mid);                 if (cur-pre >= n)                    rr = mid;                else                    ll = mid;            }            if (f (ll)-pre >= n) {                printf ("%lld\n", ll);            }            else if (f (rr)-pre >= n) {                printf ("%lld\n", rr);            }            else                printf ("Nya!\n");        }    }    return 0;}