牛客练习赛29—F算式子

题目链接：传送门

 

题解：

 

代码：

#include 
#include 
#include 
#include 

using namespace std;

typedef long long llt;

const int N = 2000010;
const int INF = 0x3fffffff;

llt cnt1[N],cnt2[N];

int main()
{
    int n,m,a;
    cin >> n >> m;
    for(int i = 0; i < n; ++i){
        cin >> a;
        cnt1[a]++;
    }
    for(int i = 1; i <= m; ++i){
        if(cnt1[i] == 0) continue;
        for(int j = i; j <= m; j += i){
            cnt2[j] += cnt1[i];
        }
    }
    for(int i = 1; i <= m; ++i)
        cnt2[i] += cnt2[i-1];
    for(int i = 1; i <= m; ++i)
        cnt1[i] += cnt1[i-1];
    llt val,ans = 0;
    for(int i = 1; i <= m; ++i){
        val = 0;
        for(int j = i; j <= m; j += i){
            int t = min(j+i-1, m);
            val += (cnt1[t]-cnt1[j-1])*(j/i);
        }
        //cout << val << endl;
        ans ^= val+cnt2[i];
    }
    cout << ans << endl;
    return 0;
}