UESTC - 1041 Hug the princess（位运算&转换）好题

UESTC - 1041

Hug the princess

Time Limit:                                                        1000MS                        Memory Limit: 65535KB 64bit IO Format:                            %lld & %llu                       

SubmitStatus

Description

There is a sequence with n elements. Assuming they are a1,a2,⋯,an.

Please calculate the following expession.

∑1≤i<j≤n(ai∧aj)+(ai|aj)+(ai&aj)

In the expression above, ^|& is bit operation. If you don’t know bit operation, you can visit

http://en.wikipedia.org/wiki/Bitwise_operation

to get some useful information.

Input

The first line contains a single integer n, which is the size of the sequence.

The second line contains n integers, the ith integer ai is the ith element of the sequence.

1≤n≤100000,0≤ai≤100000000

Output

Print the answer in one line.

Sample Input

2
1 2

Sample Output

6

Hint

Because the answer is so large, please use long long instead of int. Correspondingly, please use%lld instead of %d to scanf and printf.

Large input. You may get Time Limit Exceeded if you use “cin” to get the input. So “scanf” is suggested.

Likewise, you are supposed to use “printf” instead of “cout”.

Source

The 13th UESTC Programming Contest Preliminary

//题意：

给你n个数，让求∑1≤i<j≤n(ai∧aj)+(ai|aj)+(ai&aj)；

//思路：

队友的想法太机智了，先用个二维数组存放每个位上的值的和，然后在每个位进行计算。

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int MAXN = 1e5 + 100;int num[MAXN][35];int a[35];int p[MAXN];typedef long long LL;int main(){int n;while(~scanf("%d", &n)){int x;memset(num, 0, sizeof(num));for(int i = 1; i <= n; i++){scanf("%d", &x);for(int j = 0; j < 33; j++){num[i][j] = num[i - 1][j] + x % 2;x = x / 2;}}LL ans = 0;for(int i = 1; i <= n; i++){for(int j = 0; j < 33; j++){a[j] = num[i][j] - num[i - 1][j];}for(int j = 0; j < 33; j++){int cnt = 0;if(a[j]){cnt += (i - 1 - num[i - 1][j]);//异或值 cnt += i - 1;   //或值 cnt += num[i - 1][j];   //非值 }else{cnt += num[i - 1][j];cnt += num[i - 1][j];}ans += (LL)cnt * (1 << j);//得到对应位上的值后再向右移动对应的位数 }}printf("%lld\n", ans);}return 0;}