【HDU】5637

Problem Description
A list of n integers are given. For an integer x you can do the following operations:

+ let the binary representation of x be b31b30...b0¯¯¯¯¯¯¯¯¯¯¯¯¯¯, you can flip one of the bits.
+ let y be an integer in the list, you can change x to x⊕y, where ⊕ means bitwise exclusive or operation.

There are several integer pairs (S,T). For each pair, you need to answer the minimum operations needed to change S to T.
 

Input
There are multiple test cases. The first line of input contains an integer T (T≤20), indicating the number of test cases. For each test case:

The first line contains two integer n and m (1≤n≤15,1≤m≤105) -- the number of integers given and the number of queries. The next line contains n integers a1,a2,...,an (1≤ai≤105), separated by a space.

In the next m lines, each contains two integers si and ti (1≤si,ti≤105), denoting a query.
 

Output
For each test cases, output an integer S=(∑i=1mi⋅zi) mod (109+7), where zi is the answer for i-th query.
 

Sample Input
1
3 3
1 2 3
3 4
1 2
3 9
 

Sample Output
10
Hint

$3 \to 4$ (2 operations): $3 \to 7 \to 4$

$1 \to 2$ (1 operation): $1 \oplus 3 = 2$

$3 \to 9$ (2 operations): $3 \to 1 \to 9$
 
 

Source
BestCoder Round #74 (div.2)
 

Recommend
wange2014

题意分析:

给出n个数的序列a,对于一个整数x,有两种操作: 
1.改变x二进制中任一位 
2.将x变为x^a[i],1<=i<=n 
m次查询,每次查询输入两个整数x和y,问x最少经过多少次操作可以变成y 

记得mod

假设 s^x^y^z^w^...^q=t 是最小操作次数,由于其等价于 0^x^y^z^w^...^q=s^t,

因此只需要根据所给的 n 个数将 1E5 范围内的所有步数求出来存到 res[] 数组中,

最后根据 s、t 的值直接可以得到 res[s^t] 然后进行计算即可。

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define PI acos(-1.0)
#define E 1e-6
#define MOD 1000000007
#define INF 0x3f3f3f3f
#define N 300001
#define LL long long
using namespace std;
int a[N],res[N];
bool vis[N];
void bfs(int n){
    memset(vis,0,sizeof(vis));
    for(int i=1;i<=2e5;i*=2)//在给出的a[i]基础上向后补1、2、4、8...
        a[n++]=i;
 
    queue Q;
    Q.push(0);
    res[0]=0;
    vis[0]=true;
    while(!Q.empty()){
        int top=Q.front();
        Q.pop();
 
        for(int i=0;i

 

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