HDU 5676 ztr loves lucky numbers - dblank

HDU 5676 ztr loves lucky numbers

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
题意:问最小的大于等于n的数,并且这个数只由4和7组成,且4和7的个数一样。
思路:直接将1~16位的所有数通过dfs打表得到,奇数位的n的答案肯定是位数加1的合法的数,但是当n大于16位最大的合法数的时候,答案是爆long long,通过特判输出。
#include<iostream>  
#include<algorithm>  
#include<cstdio>  
#include<queue>  
#include<map>  
#include<vector>  
#include<cstring>  
#include<cmath>  
using namespace std;  
typedef long long ll;  
const int inf =0x3f3f3f3f;  
const double  pi = acos(-1.0);  
const int N = 1e5 + 10;  
ll s[N*5];  
int k;  
void dfs(ll num, int a, int b)  
{  
    if(a+b == 19)  
        return ;  
    if(a == b && a+b)  
        s[k++] = num;  
    dfs(num*10 + 4, a+1, b);  
    dfs(num*10 + 7, a, b+1);  
}  
int main()  
{  
    int t;  
    k = 0;  
    dfs(0, 0, 0);  
    sort(s, s+k);  
    cin>>t;  
    while(t--)  
    {  
        ll n;  
        scanf("%I64d", &n);  
        int index = lower_bound(s, s+k, n) - s;  
        if(index == k)  
            printf("44444444447777777777\n");  
        else printf("%I64d\n", s[sb]);  
    }  
    return 0;  
}

 

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