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 9Sample 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<iostream>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<cstring>
#include<cmath>
#include<ctime>
#include<algorithm>
#include<utility>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<map>
#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<int> Q;Q.push(0);res[0]=0;vis[0]=true;while(!Q.empty()){int top=Q.front();Q.pop();for(int i=0;i<n;i++){if(vis[a[i]^top]==false){Q.push(a[i]^top);//相应异或元素进队res[a[i]^top]=res[top]+1;//步数+1vis[a[i]^top]=true;//标记}}}
}
int main()
{int t;scanf("%d",&t);while(t--){int n,m;scanf("%d%d",&n,&m);for(int i=0;i<n;i++)scanf("%d",&a[i]);bfs(n);LL sum=0;int s,t;for(int i=1;i<=m;i++){scanf("%d%d",&s,&t);sum=(sum+(LL)i*res[s^t])%MOD;}printf("%d\n",(int)sum);}return 0;
}