CDOJ 1041 Hug the princess 【思维】

思维】"/>

CDOJ 1041 Hug the princess 【思维】

题目链接：CDOJ 1041 Hug the princess

题意：求解 ∑ni=1 $\sum_{i=1}^n$ a[i] ^ a[j] + a[i] | a[j] + a[i] & a[j]。

思路：a ^ b + a | b + a & b = a + b + a | b - a & b。先统计a + b的贡献，而a | b + a & b的贡献则是每个二进制位上1的个数，最后累加一下就好了。

AC代码：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <queue>
#include <set>
#define CLR(a, b) memset(a, (b), sizeof(a))
#define fi first
#define se second
using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
const int MAXN = 1e5 +10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
void add(LL &x, LL y) { x += y; x %= MOD; }
LL a[MAXN], sum[MAXN], bit[MAXN][41], sbit[41];
int main()
{int n;while(scanf("%d", &n) != EOF) {sum[0] = 0;for(int j = 0; j <= 40; j++) {bit[0][j] = 0; sbit[j] = 0;}for(int i = 1; i <= n; i++) {scanf("%lld", &a[i]);sum[i] = sum[i-1] + a[i];for(int j = 0; j <= 40; j++) {bit[i][j] = 0;if(a[i] & (1LL << j)) {bit[i][j] = 1;}bit[i][j] += bit[i-1][j];}}LL ans = 0;for(int i = 1; i < n; i++) {ans += (n - i) * a[i] + sum[n] - sum[i];for(int j = 0; j <= 40; j++) {int num = bit[n][j] - bit[i][j];if(a[i] & (1LL << j)) {sbit[j] += n - num - i;}else {sbit[j] += num;}}}LL res = 0;for(int i = 0; i <= 40; i++) {res += sbit[i] * (1LL << i);}printf("%lld\n", ans + res);}return 0;
}