Counting Pair-白红宇

Counting Pair

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发布时间：2019-06-14

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Counting Pair

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

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Best Solution

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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

4 5

Sample Output

Case #1:

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

1 #include 
       
         2 #include 
        
          3 #include 
         
           4 #include 
           5 #include 
           
             6 #include 
            
              7 #include 
             
               8 #include 
              
                9 #include 
               
                10 11 using namespace std;12 13 int T, N, M, Q, s, cnt;14 15 int main()16 {17 scanf("%d", &T);18 for(int ca = 1; ca <= T; ca++)19 {20 scanf("%d %d", &N, &M);21 scanf("%d", &Q);22 printf("Case #%d:\n", ca);23 while(Q--)24 {25 scanf("%d", &s);26 if(s <= 1 || s > M + N) cnt = 0;27 else28 {29 if(N < M) swap(M, N);30 if(s <= M) cnt = s - 1;31 else if(s > M && s <= N) cnt = M;32 else if(s > N) cnt = M + N - s + 1;33 }34 printf("%d\n",cnt);35 }36 }37 return 0;38 }