ztr loves lucky numbers
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1834 Accepted Submission(s): 707
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.
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.
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
Output the answer
Sample Input
2
4500
47
Sample Output
4747
47
Source
思路：数总共没多少，打表即可，再最大的情况特判一下；
#pragma comment(linker, "/STACK:1024000000,1024000000") #include<iostream> #include<cstdio> #include<cmath> #include<string> #include<queue> #include<algorithm> #include<stack> #include<cstring> #include<vector> #include<list> #include<set> #include<map> using namespace std; #define ll long long #define pi (4*atan(1.0)) #define eps 1e-14 #define bug(x) cout<<"bug"<<x<<endl; const int N=1e2+10,M=1e6+10,inf=1e9+10; const ll INF=5e17+10,mod=1e9+7; ///数组大小 int check(ll x) { int s=0,q=0; while(x) { if(x%10==4)s++; else q++; x/=10; } if(s==q)return 1; return 0; } vector<ll>ans; vector<ll>::iterator it; void dfs(ll x) { if(x>INF)return; if(check(x))ans.push_back(x); dfs(x*10+4); dfs(x*10+7); } int main() { dfs(4),dfs(7); sort(ans.begin(),ans.end()); int T; scanf("%d",&T); while(T--) { ll x; scanf("%lld",&x); it=lower_bound(ans.begin(),ans.end(),x); if(it==ans.end()) printf("44444444447777777777 "); else printf("%lld ",*it); } return 0; }