CDOJ 1041 Hug the princess 【思维】

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题目链接：CDOJ 1041 Hug the princess

题意：求解∑ni=1 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 secondusing 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;}

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