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 = 1; ca <= T; ca++)
    {
        scanf("%d %d", &N, &M);
        scanf("%d", &Q);
        printf("Case #%d:\n", ca);
        while(Q--)
        {
            scanf("%d", &s);
            if(s <= 1 || s > M + N) cnt = 0;
            else
            {
                if(N < M) swap(M, N);
                if(s <= M) cnt = s - 1;
                else if(s > M && s <= N) cnt = M;
                else if(s > N) cnt = M + N - s + 1;
            }
            printf("%d\n",cnt);
        }
    }
    return 0;
}