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(1≤n≤105) cases

For each cases:

The only line contains a positive integer n(1≤n≤1018). This number doesn't have leading zeroes.

 

Output

For each cases
Output the answer

 

Sample Input

2450047

 

Sample Output

474747

 

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 <map>#include <set>#include <queue>#include <cmath>#include <stack>#include <cctype>#include <cstdio>#include <cstring>#include <stdlib.h>#include <iostream>#include <algorithm>#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-8ll 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;    queue<long long>q;    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(n<INF){            m=n*10+4;            q.push(m);            m=n*10+7;            q.push(m);        }    }}int main(){    bfs();    sort(dp,dp+tot);    //cout<<tot<<'\12';    int t;    ll a;    scanf("%d",&t);    while(t--){        scanf("%I64d",&a);        if(a>dp[tot-1])printf("44444444447777777777\n");        else printf("%I64d\n",dp[lower_bound(dp,dp+tot,a)-dp]);    }    return 0;}