并查集 + 位运输

先用并查集处理每个联通块，并把权值排序。

然后对于 a * (c & d) 可以把c和d才成二进制，同样位置上存在1才对答案有贡献，所以我们对每个联通块，枚举每个位置是1的数量，算出贡献就好了。

#include 
    
     #define INF 0x3f3f3f3f#define full(a, b) memset(a, b, sizeof a)#define FAST_IO ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)using namespace std;typedef long long ll;inline int lowbit(int x){ return x & (-x); }inline int read(){    int ret = 0, w = 0; char ch = 0;    while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }    while(isdigit(ch)) ret = (ret << 3) + (ret << 1) + (ch ^ 48), ch = getchar();    return w ? -ret : ret;}inline int gcd(int a, int b){ return b ? gcd(b, a % b) : a; }inline int lcm(int a, int b){ return a / gcd(a, b) * b; }template 
     
      inline T max(T x, T y, T z){ return max(max(x, y), z); }template 
      
       inline T min(T x, T y, T z){ return min(min(x, y), z); }template 
       
        inline A fpow(A x, B p, C lyd){    A ans = 1;    for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;    return ans;}const int N = 100005;const int MOD = 1e9 + 7;int _, n, m, v[N], parent[N], t[30];vector
        
          vi[N];int find(int p){    while(p != parent[p]){        parent[p] = parent[parent[p]];        p = parent[p];    }    return p;}bool isConnect(int p, int q){    return find(p) == find(q);}void merge(int p, int q){    int pRoot = find(p), qRoot = find(q);    if(pRoot == qRoot) return;    parent[qRoot] = pRoot;}void build(){    for(int i = 0; i <= n; i ++){        parent[i] = i;        vi[i].clear();    }}int main(){    for(_ = read(); _; _ --){        n = read(), m = read(), build();        for(int i = 1; i <= n; i ++) v[i] = read();        for(int i = 1; i <= m; i ++){            int x = read(), y = read();            if(isConnect(x, y)) continue;            merge(x, y);        }        for(int i = 1; i <= n; i ++) vi[find(i)].push_back(v[i]);        for(int i = 1; i <= n; i ++) sort(vi[i].begin(), vi[i].end());        ll ans = 0;        for(int i = 1; i <= n; i ++){            if(vi[i].size() <= 1) continue;            full(t, 0);            for(auto val: vi[i]){                for(int j = 0; j < 32; j ++){                    if(val & (1 << j)){                        ans = (ans % MOD + 1LL * t[j] * val % MOD * (1 << j) % MOD) % MOD;                        t[j] ++;                    }                }            }        }        printf("%lld\n", ans);    }    return 0;}