HDU-3711 Binary Number(贪心)

HDU-3711 Binary Number(贪心)

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

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

题意：有A、B两个集合，B集合每个元素都能在A集合里面找到使得f(ai,bi)【两数异或之后的二进制 1的个数】最小的一个元素ai，直接贪心。

#include <algorithm>#include <iostream>#include <cstring>#include <cstdio>#include <ctime>#include <set>#include <map>#include <cmath>using namespace std;#define pi 3.14159265359typedef long long LL;const int maxn = 110;int n, t, m, ma, maa;int a[maxn],b[maxn];int b_f(int x){    int sum = 0;    while(x){        sum += x%2;        x/=2;    }    return sum;}int main(){    scanf("%d",&t);    while(t--){        scanf("%d %d",&n,&m);        for(int i=1;i<=n;i++){            scanf("%d",&a[i]);        }        for(int i=1;i<=m;i++){            scanf("%d",&b[i]);            ma = 99999999;            maa = a[1];            for(int j=1;j<=n;j++){                int temp = b_f( a[j]^b[i] );                if(ma > temp){                    ma = temp;                    maa = a[j];                }                if(ma==temp){                    maa = a[j] < maa ? a[j] : maa;                }               }            printf("%d\n",maa);        }    }    return 0;}