HDU 5676 ztr loves lucky numbers

—亚信科技，巴卡斯（杭州），壹晨仟阳(杭州)，英雄互娱（杭州）  （包括2016级新生）除了校赛，还有什么途径可以申请加入ACM校队？

ztr loves lucky numbers Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 51    Accepted Submission(s): 24 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 2 4500 47   Sample Output 4747 47 把1到10^20中满足条件的全部枚举出来，然后再二分查找就可以了。 注意20位最小的44444444447777777777在是超long long int 的 只好进行一下特判 #include <iostream> #include <string.h> #include <stdlib.h> #include <algorithm> #include <math.h> #include <stdio.h> #include <string> #include <map> using namespace std; #define MAX 600000 typedef __int64 LL; LL n; LL a[MAX]; int cnt; void dfs(LL x,LL y,LL num) { if(x==0&&y==0) { a[++cnt]=num; return; } if(x>0) dfs(x-1,y,num*10+4); if(y>0) dfs(x,y-1,num*10+7); } void fun() { a[1]=47; a[2]=74; cnt=2; for(int i=4;i<=18;i+=2) { dfs(i/2,i/2,0); } } int main() { int t; scanf("%d",&t); fun(); sort(a+1,a+cnt+1); while(t--) { scanf("%I64d",&n); if(n>777777777444444444) { cout<<"44444444447777777777"<<endl; continue; } LL l=1,r=cnt; LL res; while(l<=r) { LL mid=(l+r)/2; if(a[mid]>=n) { res=a[mid]; r=mid-1; } else l=mid+1; } printf("%I64d\n",res); } return 0; }