sdau 省赛热身4 C - Counting Pair

sdau 省赛热身4   C - Counting Pair      

C - Counting Pair

Time Limit:1000MS     Memory Limit:65535KB     64bit IO Format:%lld & %llu

Submit Status

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.

数字组合

题目大意：一个party来了n男m女，每人有其对应（1~n）或（1~m）编号，主人给出一数，若男女两队人马中有两数之和等于此数，则可配对，问给出一数，可配多少对。

题目分析：若给出的数x既小于m，又小于n，则x-1就是答案。eg.在本例第三个数据中3可配1~2和2~1两对。但如果给出的数-1大于m，n之一，就出现问题了，组合往上找的时候会meet上界，这时需要把上界之上的部分减掉。（这部分代码挺绕的，不过多看两眼肯定能明白）

code:

#include<stdio.h>int main(){int i,j,t,n,m,p,count,k,l,test;scanf("%d",&t);for(i=1;i<=t;i++){scanf("%d%d%d",&n,&m,&p);printf("Case #%d:\n",i);for(j=0;j<p;j++){count=0;scanf("%d",&test);/*test--;for(k=1;k<=n;k++){if(test-k>0&&test-k<=m)count++;else if(test-k<=0)break;}printf("%d\n",count);*/if(test-1<=n&&test-1<=m)printf("%d\n",test-1);else if(test>n+m)printf("0\n");else printf("%d\n",test-1-(test>n?test-1-n:0)-(test>m?test-1-m:0));}}}

PS：今天状态尚可，A出三道，后面还有一道。（continue praying）