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 程式師世界 >> 編程語言 >> C語言 >> C++ >> C++入門知識 >> 杭電 3711 Binary Number

杭電 3711 Binary Number

編輯:C++入門知識

Binary Number

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1287 Accepted Submission(s): 807


Problem Description For 2 non-negative integers x and y, f(x, y) is defined as the number of different bits in the binary format of x and y. For example, f(2, 3)=1,f(0, 3)=2, f(5, 10)=4. Now given 2 sets of non-negative integers A and B, for each integer b in B, you should find an integer a in A such that f(a, b) is minimized. If there are more than one such integer in set A, choose the smallest one.
Input The first line of the input is an integer T (0 < T ≤ 100), indicating the number of test cases. The first line of each test case contains 2 positive integers m and n (0 < m, n ≤ 100), indicating the numbers of integers of the 2 sets A and B, respectively. Then follow (m + n) lines, each of which contains a non-negative integers no larger than 1000000. The first m lines are the integers in set A and the other n lines are the integers in set B.
Output For each test case you should output n lines, each of which contains the result for each query in a single line.
Sample Input
2
2 5
1
2
1
2
3
4
5
5 2
1000000
9999
1423
3421
0
13245
353

Sample Output
1
2
1
1
1
9999
0

AC代碼如下:

#include 
int a[105];
int count(int x)
{
    int c = 0;
    for(;x;x>>=1) if(x&1) c++;
    return c;
}
int main()
{
    int b, i, j, n, m, k, min, t,cases;
    scanf("%d",&cases);
    while(cases--)
    {
        scanf("%d%d",&n,&m);
        for(i=0; i

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