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| Time Limit: 1000MS | Memory Limit: 65535KB | 64bit IO Format: %lld & %llu |
Description
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
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
#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];
}
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;
}UESTC--1041--Hug the princess（位运算）
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原文地址：http://blog.csdn.net/qq_29963431/article/details/51346098
