Hug the princess（思维，位运算）

Hug the princess

Time Limit: 3000/1000MS (Java/Others) Memory Limit: 65535/65535KB (Java/Others)

There is a sequence with

Please calculate the following expession.

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

The second line contains

Output

Print the answer in one line.

Sample input and output

Sample Input	Sample Output
2 1 2	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".

题解：

让求这个表达式，我们可以把求出每一位的前缀和，然后再根据每一位找出前面是1，是0的当前位的个数；然后加上cnt * (1 << j)就好了

代码：

#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++){
            int temp = 0, pos = -1;
            for(int j = 0; j < 33; j++){
                a[j] = num[i][j] - num[i - 1][j];
            //    printf("%d ", a[j]);
            }//puts("");
            for(int j = 0; j < 33; j++){
                int cnt = 0;
                if(a[j]){
                    cnt += (i - 1 - num[i - 1][j]);
                    cnt += i - 1;
                    cnt += num[i - 1][j];
                }
                else{
                    cnt += num[i - 1][j];
                    cnt += num[i - 1][j];
                }
                ans += (LL)cnt * (1 << j);
            }
        }
        printf("%lld
", ans);
    }
    return 0;
}