hdu 5676

ztr loves lucky numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 530    Accepted Submission(s): 224


Problem Description
ztr loves lucky numbers. Everybody knows that positive integers are lucky if their decimal representation doesn't contain digits other than 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.

Lucky number is super lucky if it's decimal representation contains equal amount of digits 4 and 7. For example, numbers 47, 7744, 474477 are super lucky and 4, 744, 467 are not.

One day ztr came across a positive integer n. Help him to find the least super lucky number which is not less than n.
 

Input
There are T (1n105) cases

For each cases:

The only line contains a positive integer  n(1n1018). This number doesn't have leading zeroes.
 

Output
For each cases
Output the answer
 

Sample Input
 
   
2 4500 47
 

Sample Output
 
   
4747 47
 

Source
BestCoder Round #82 (div.2)
 

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accode:
#pragma warning(disable:4786)//使命名长度不受限制
#pragma comment(linker, "/STACK:102400000,102400000")//手工开栈
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define rd(x) scanf("%d",&x)
#define rd2(x,y) scanf("%d%d",&x,&y)
#define rd3(x,y,z) scanf("%d%d%d,&x,&y,&z)
#define rdl(x) scanf("%I64d,&x);
#define rds(x) scanf("%s",x)
#define rdc(x) scanf("%c",&x)
#define ll long long int
#define ull unsigned long long
#define maxn 9999999
#define mod 1000000007
#define INF 77777777777777778LL
#define FOR(i,f_start,f_end) for(int i=f_start;i<=f_end;++i)
#define MT(x,i) memset(x,i,sizeof(x))
#define PI  acos(-1.0)
#define E  exp(1)
#define eps 1e-8
ll gcd(ll a,ll b){return b==0?a:gcd(b,a%b);}
ll mul(ll a,ll b,ll p){ll sum=0;for(;b;a=(a+a)%p,b>>=1)if(b&1)sum=(sum+a)%p;return sum;}
inline void Scan(int &x) {
      char c;while((c=getchar())<'0' || c>'9');x=c-'0';
      while((c=getchar())>='0' && c<='9') x=(x<<3)+(x<<1)+c-'0';
}
using namespace std;
ll dp[maxn];
int tot=0;
void bfs(){
    ll n,t,m;
    queueq;
    q.push(4LL);
    q.push(7LL);
    tot =0;
    while(!q.empty()){
        n=q.front();q.pop();
        t=n;
        int x,y;
        x=y=0;
        while(t){
            if(t%10==4)x++;
            else y++;
            t/=10;
        }
        if(x==y)
            dp[tot++]=n;
        if(ndp[tot-1])printf("44444444447777777777\n");
        else printf("%I64d\n",dp[lower_bound(dp,dp+tot,a)-dp]);
    }
    return 0;
}


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