HDU 3943 K-th Nya Number（数位dp+二分） - c++编程基础

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HDU 3943 K-th Nya Number（数位dp+二分）

2015-11-21 01:04:32 来源: 作者: 【大中小】浏览:1次

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 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

Sample Output

Case #1:
47
74
147
174
247
274
347
374
Nya!
Nya!

Author hzhua

#include
   
    
#include
    
      #include
     
       #include
      
        #include
       
         #include
        
          #include
         
           #include
          
            #include
           
             #include
             #define L(x) (x<<1) #define R(x) (x<<1|1) #define MID(x,y) ((x+y)>>1) #define eps 1e-8 typedef __int64 ll; #define fre(i,a,b) for(i = a; i 
             
              = a;i--) #define mem(t, v) memset ((t) , v, sizeof(t)) #define ssf(n) scanf("%s", n) #define sf(n) scanf("%d", &n) #define sff(a,b) scanf("%d %d", &a, &b) #define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c) #define pf printf #define bug pf("Hi\n") using namespace std; #define INF 0x3f3f3f3f #define N 25 ll dp[N][N][N]; int bit[N]; ll le,ri; int x,y,k; ll dfs(int pos,int prex,int prey,int bound) { if(pos==0) return prex==0&&prey==0; if(prex<0||prey<0) return 0; if(!bound&&dp[pos][prex][prey]!=-1) return dp[pos][prex][prey]; int up=bound ? bit[pos]:9; ll temp=0; int i; fre(i,0,up+1) { temp+=dfs(pos-1,prex-(i==4),prey-(i==7),bound&&i==up); } if(!bound) dp[pos][prex][prey]=temp; return temp; } ll fdd(ll n) { int len=0; while(n) { bit[++len]=n%10; n/=10; } return dfs(len,x,y,1); } void solve() { ll prex=fdd(le); ll prey=fdd(ri); ll lle=le,rri=ri; ll mid,ans; k+=prex; while(lle<=rri) { mid=(lle+rri)>>1; if(fdd(mid)>=k) { ans=mid; rri=mid-1; } else lle=mid+1; } pf("%I64d\n",ans); } int main() { int i,j,t,ca=0; sf(t); mem(dp,-1); while(t--) { pf("Case #%d:\n",++ca); scanf("%I64d%I64d",&le,&ri); sff(x,y); ll numle=fdd(le); ll numri=fdd(ri); //pf("%I64d %I64d\n",numle,numri); int m; sf(m); while(m--) { sf(k); if(numle+k>numri) puts("Nya!"); else solve(); } } return 0; }


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