Counting Pair

Counting Pair

Time Limit: 1000 ms Memory Limit: 65535 kB Solved: 112 Tried: 1209

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Description

 

Bob hosts a party and invites N boys and M girls. He gives every boy here a unique number Ni(1 <= Ni <= N). And for the girl, everyone holds a unique number Mi(1 <= Mi <= M), too.

Now when Bob name a number X, if a boy and a girl wants and their numbers' sum equals to X, they can get in pair and dance.

At this night, Bob will name Q numbers, and wants to know the maxinum pairs could dance in each time. Can you help him?

 

Input

 

First line of the input is a single integer T(1 <= T <= 30), indicating there are T test cases.

The first line of each test case contains two numbers N and M(1 <= N,M <= 100000).

The second line contains a single number Q(1 <= Q <= 100000).

Each of the next Q lines contains one number X(0 <= X <= 10^9), indicating the number Bob names.

 

Output

 

For each test case, print "Case #t:" first, in which t is the number of the test case starting from 1.

Then for each number Bob names, output a single num in each line, which shows the maxinum pairs that could dance together.

 

Sample Input

 

1
4 5
3
1
2
3

 

Sample Output

 

Case #1:
0
1
2

 

Hint

 

This problem has very large input data. scanf and printf are recommended for C++ I/O.

 

Source

 

Sichuan State Programming Contest 2012

 看代码就懂了

 #include <iostream>
 #include <queue>
 #include <vector>
 #include <map>
 #include <string>
 #include <algorithm>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 
 using namespace std;
 
 int T, N, M, Q, s, cnt;
 
 int main()
 {
     scanf("%d", &T);
     for(int ca = ; ca <= T; ca++)
     {
         scanf("%d %d", &N, &M);
         scanf("%d", &Q);
         printf("Case #%d:\n", ca);
         while(Q--)
         {
             scanf("%d", &s);
             if(s <=  || s > M + N) cnt = ;
             else
             {
                 if(N < M) swap(M, N);
                 if(s <= M) cnt = s - ;
                 else if(s > M && s <= N) cnt = M;
                 else if(s > N) cnt = M + N - s + ;
             }
             printf("%d\n",cnt);
         }
     }
     return ;
 }