Counting Pair

Counting Pair

Time Limit:1000 msMemory Limit:65535 kBSolved:112Tried: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

1 #include <iostream>
 2 #include <queue>
 3 #include <vector>
 4 #include <map>
 5 #include <string>
 6 #include <algorithm>
 7 #include <cstdio>
 8 #include <cstring>
 9 #include <cmath>
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             else
28             {
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 }

原文链接: https://www.cnblogs.com/cszlg/p/3220713.html

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