Hug the princess（思维，位运算）-优快云博客

本文介绍了一道关于位运算的编程题目，需要计算特定位运算表达式的总和。通过预处理每一位的前缀和，并统计每一位上1和0的数量，最终实现了高效求解。

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Hug the princess

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

Submit Status

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\n", ans);
    }
    return 0;
}