Codeforces Round #383 (Div. 2) Arpa’s obvious problem and Mehrdad’s terrible solution 数学
2017-05-07 00:19
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题目:
There are some beautiful girls in Arpa’s land as mentioned before.
Once Arpa came up with an obvious problem:
Given an array and a number x, count the number of pairs of indices i, j (1 ≤ i < j ≤ n)
such that
![](http://codeforces.com/predownloaded/f0/93/f0934fb9630fb6804ec22619326ce588c6c332d7.png)
,
where
![](http://codeforces.com/predownloaded/6b/93/6b937d7fb50dfbb98833cf050e078dca41a13c0f.png)
is
bitwise xor operation (see notes for explanation).
![](http://codeforces.com/predownloaded/7a/c1/7ac107ccd6dd880f227dd2a9ce9742ed265d2184.png)
Immediately, Mehrdad discovered a terrible solution that nobody trusted. Now Arpa needs your help to implement the solution to that problem.
Input
First line contains two integers n and x (1 ≤ n ≤ 105, 0 ≤ x ≤ 105) —
the number of elements in the array and the integer x.
Second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 105) —
the elements of the array.
Output
Print a single integer: the answer to the problem.
Examples
input
output
input
output
Note
In the first sample there is only one pair of i = 1 and j = 2.
![](http://codeforces.com/predownloaded/68/ac/68ace19cfe8062d209d7233236619123b6146e63.png)
so
the answer is 1.
In the second sample the only two pairs are i = 3, j = 4 (since
![](http://codeforces.com/predownloaded/c5/67/c5679f470aa199c954d8f147670732be255009b5.png)
)
and i = 1, j = 5 (since
![](http://codeforces.com/predownloaded/55/1d/551d516fd74e0f3774ecdb9d19168383845eb37d.png)
).
A bitwise xor takes two bit integers of equal length and performs the logical xor operation
on each pair of corresponding bits. The result in each position is 1 if only the first bit is 1 or
only the second bit is 1, but will be0 if both are 0 or
both are 1. You can read more about bitwise xor operation here:https://en.wikipedia.org/wiki/Bitwise_operation#XOR.
就是给你一个序列,问有多少个二元组(a,b)满足a^b=c(c给定).
1,异或有一个性质 a^b^a=b.那么a^b=c --> a^b^a=c^a -->a^c=b
2,a^b最大值可以近似取到max(a,b)*2 例如 011^100=111开数组时不注意会RE的
3,int*int可能会返回一个负数,也就是溢出。如a=100000,res=a*a, 那么res=-727379968。
4,这种题主要是是考虑重复,已经使用过的二元组用visit数组标记,那么问题来了,如果x=0,序列是1 1 1 1时该怎么办,我们不难发现这个就是cnt[i]*(cnt[i]-1)/2(这个地方可能会发生第三点的情况)
code:
#include<cstdio>
const int MAXN=2e5+5;
int a[MAXN];
long long cnt[MAXN];
long long visit[MAXN];
int main(){
int n,x;scanf("%d%d",&n,&x);
for(int i=0;i<n;++i){
scanf("%d",a+i);
cnt[a[i]]++;
}
long long res=0;
for(int i=1;i<=100000;++i){
if(!visit[i]&&cnt[i]&&cnt[x^i]){
if(i==(x^i))res+=cnt[i]*(cnt[i]-1)/2;
else res+=cnt[i]*cnt[x^i];
visit[i]=visit[x^i]=1;
}
}
printf("%I64d\n",res);
}
还要继续加油呀!
There are some beautiful girls in Arpa’s land as mentioned before.
Once Arpa came up with an obvious problem:
Given an array and a number x, count the number of pairs of indices i, j (1 ≤ i < j ≤ n)
such that
![](http://codeforces.com/predownloaded/f0/93/f0934fb9630fb6804ec22619326ce588c6c332d7.png)
,
where
![](http://codeforces.com/predownloaded/6b/93/6b937d7fb50dfbb98833cf050e078dca41a13c0f.png)
is
bitwise xor operation (see notes for explanation).
![](http://codeforces.com/predownloaded/7a/c1/7ac107ccd6dd880f227dd2a9ce9742ed265d2184.png)
Immediately, Mehrdad discovered a terrible solution that nobody trusted. Now Arpa needs your help to implement the solution to that problem.
Input
First line contains two integers n and x (1 ≤ n ≤ 105, 0 ≤ x ≤ 105) —
the number of elements in the array and the integer x.
Second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 105) —
the elements of the array.
Output
Print a single integer: the answer to the problem.
Examples
input
2 3 1 2
output
1
input
6 1 5 1 2 3 4 1
output
2
Note
In the first sample there is only one pair of i = 1 and j = 2.
![](http://codeforces.com/predownloaded/68/ac/68ace19cfe8062d209d7233236619123b6146e63.png)
so
the answer is 1.
In the second sample the only two pairs are i = 3, j = 4 (since
![](http://codeforces.com/predownloaded/c5/67/c5679f470aa199c954d8f147670732be255009b5.png)
)
and i = 1, j = 5 (since
![](http://codeforces.com/predownloaded/55/1d/551d516fd74e0f3774ecdb9d19168383845eb37d.png)
).
A bitwise xor takes two bit integers of equal length and performs the logical xor operation
on each pair of corresponding bits. The result in each position is 1 if only the first bit is 1 or
only the second bit is 1, but will be0 if both are 0 or
both are 1. You can read more about bitwise xor operation here:https://en.wikipedia.org/wiki/Bitwise_operation#XOR.
就是给你一个序列,问有多少个二元组(a,b)满足a^b=c(c给定).
1,异或有一个性质 a^b^a=b.那么a^b=c --> a^b^a=c^a -->a^c=b
2,a^b最大值可以近似取到max(a,b)*2 例如 011^100=111开数组时不注意会RE的
3,int*int可能会返回一个负数,也就是溢出。如a=100000,res=a*a, 那么res=-727379968。
4,这种题主要是是考虑重复,已经使用过的二元组用visit数组标记,那么问题来了,如果x=0,序列是1 1 1 1时该怎么办,我们不难发现这个就是cnt[i]*(cnt[i]-1)/2(这个地方可能会发生第三点的情况)
code:
#include<cstdio>
const int MAXN=2e5+5;
int a[MAXN];
long long cnt[MAXN];
long long visit[MAXN];
int main(){
int n,x;scanf("%d%d",&n,&x);
for(int i=0;i<n;++i){
scanf("%d",a+i);
cnt[a[i]]++;
}
long long res=0;
for(int i=1;i<=100000;++i){
if(!visit[i]&&cnt[i]&&cnt[x^i]){
if(i==(x^i))res+=cnt[i]*(cnt[i]-1)/2;
else res+=cnt[i]*cnt[x^i];
visit[i]=visit[x^i]=1;
}
}
printf("%I64d\n",res);
}
还要继续加油呀!
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