Bestcoder round # 3hd 4907

1. HDU 4907: http://acm.hdu.edu.cn/showproblem.php? PID = 1, 4907

I won't talk about Chinese questions. Let's talk about how to solve the problem!

① The first approach is to clear all array a to zero during the competition. When the input machine executes the I-th task at the Ti time, set a [Ti] to 1 and start to input Q (indicating that there is a task request beyond the worksheet in the Q time) to write a loop, so that array a starts from a [Q, until the value of a [J] is 0 and the output J is found. This method has exceeded the time limit. However, preprocessing is performed before Q is input.

Code attached:

 1 #include<stdio.h> 2 #include<string.h> 3 int t[200010],s[200010]; 4 int main() 5 { 6     int T,m,n,i,j,k,a; 7     scanf("%d",&T); 8     while(T--) 9     {10         scanf("%d%d",&m,&n);11         memset(t,0,sizeof(t));12         for(i=1; i<=m; i++)13         {14             scanf("%d",&a);15             t[a]=1;16         }17         memset(s,0,sizeof(s));18         k=1;19         for(i=1; i<=200000; i++)20             if(t[i]==0)21             {22                 for(j=k; j<=i; j++)23                     s[j]=i;24                 k=i+1;25             }26         while(n--)27         {28             scanf("%d",&a);29             printf("%d\n",s[a]);30         }31     }32     return 0;33 }

② The second approach is the binary method. Input a [] and the lower order in the back row, and record the position where each number appears. If Q does not exist, output Q directly,

If Q exists, assuming that Q is located at POs, then the binary [POs, N ], if mid-P = A [Mid]-A [p], Left = Mid + 1; otherwise right = mid-1.

Code attached:

 1 #include<iostream> 2 #include<algorithm> 3 #include<stdio.h> 4 #include<string.h> 5 using namespace std; 6 #define maxn 100010 7 int a[maxn]; 8 int main() 9 {10     int T,i;11     scanf("%d",&T);12     while(T--)13     {14         int n,m;15         scanf("%d%d",&n,&m);16         for(i=0; i<n; i++)17             scanf("%d",&a[i]);18         sort(a,a+n);19         while(m--)20         {21             int q;22             scanf("%d",&q);23             int pos=int(lower_bound(a,a+n,q)-a);24             if(pos==n||a[pos]!=q)25                 printf("%d\n",q);26             else27             {28                 int left=pos+1,right=n;29                 while(left<right)30                 {31                     int mid=(left+right)>>1;32                     if(a[mid]-q==mid-pos)33                         left=mid+1;34                     else35                         right=mid;36                 }37                  printf("%d\n",a[left-1]+1);38             }39         }40     }41     return 0;42 }

Postscript:

The lower_bound () function is in the first and lastForward and backward openReturns values greater than or equal to Val.First elementLocation. If all elements are smaller than Val, returnLast.