Codeforces Global Round 8 D - AND, OR and square sum 尽量往大的数字上移动
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#include<map> #include<queue> #include<time.h> #include<limits.h> #include<cmath> #include<ostream> #include<iterator> #include<set> #include<stack> #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; #define rep_1(i,m,n) for(int i=m;i<=n;i++) #define mem(st) memset(st,0,sizeof st) int read() { int res=0,ch,flag=0; if((ch=getchar())==‘-‘) //判断正负 flag=1; else if(ch>=‘0‘&&ch<=‘9‘) //得到完整的数 res=ch-‘0‘; while((ch=getchar())>=‘0‘&&ch<=‘9‘) res=res*10+ch-‘0‘; return flag?-res:res; } typedef long long ll; typedef pair<int,int> pii; typedef unsigned long long ull; typedef pair<double,double> pdd; const int inf = 0x3f3f3f3f; const int N = 1000100; int n; int a[N], cnt[N]; int main() { //1具有传递性,那么就把1尽量往大的数字上传 n = read(); for (int i = 1; i <= n; i ++) { a[i] = read(); for (int j = 0; j <= 20; j ++) if (a[i] >> j & 1) cnt[j] ++; } long long ans = 0; for (int i = 1, tar; i <= n; i ++) { tar = 0; for (int j = 0; j <= 20; j ++) if (cnt[j]) tar |= (1 << j), cnt[j] --; ans += 1ll * tar * tar; } printf("%lld ", ans); return 0; }
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