Codeforces Round #383 (Div. 2)B -Arpa's loud Owf and Mehrdad's evil plan
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题意:
给定n个数字 和一个x,接下来有n个数字,问数字两两组合,可以有多少情况 两个数字异或等于x
思路:
a^b=c ---> b= c^a
PS:信心满满的过题,然而把int 开成long long 了。心疼
注意0的特判
#include <iostream>#include <stdio.h>#include <cstring>#include <algorithm>#include <queue>using namespace std;typedef long long ll;long long a[1000005];long long b[1000005];int main(){ //cout<<(5^1)<<endl; int n,x; scanf("%d%d",&n,&x); for(int i=1;i<=n;i++) { int t; scanf("%d",&t); b[t]++; a[t]=1; } if(x!=0) { unsigned long long ans=0; for(int i=1;i<=1000005;i++) { if(a[i]==0) continue; int aim=(x^i); if(aim>1e+5) continue; if(a[aim]) { a[aim]=0; ans+=(b[i]*b[aim]); } } cout<<ans<<endl; } else { unsigned long long ans=0; for(int i=1;i<=1000005;i++) { if(a[i]) ans+=(b[i]-1)*(b[i])/2; } cout<<ans<<endl; } return 0;}
0 0
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