标签:and c++ force print codeforce define can main eof
思路:
满足异或值为0的区间,必须满足一下条件:
1.区间中二进制1的个数和为偶数个;
2.区间二进制1的个数最大值的两倍不超过区间和.
如果区间长度大于128,第二个条件肯定满足,所以我们只要暴力区间长度小于128的就可以了
代码:
#pragma GCC optimize(2) #pragma GCC optimize(3) #pragma GCC optimize(4) #include<bits/stdc++.h> using namespace std; #define fi first #define se second #define pi acos(-1.0) #define LL long long //#define mp make_pair #define pb push_back #define ls rt<<1, l, m #define rs rt<<1|1, m+1, r #define ULL unsigned LL #define pll pair<LL, LL> #define pii pair<int, int> #define piii pair<pii, pii> #define mem(a, b) memset(a, b, sizeof(a)) #define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0); #define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout); //head const int N = 3e5 + 100; LL a[N]; int cnt[N], sum[N][2]; int main() { int n; scanf("%d", &n); for (int i = 1; i <= n; i++) scanf("%lld", &a[i]); for (int i = 1; i <= n; i++) { for (int j = 0; j < 63; j++) { if(a[i] & (1LL << j)) cnt[i]++; } } sum[0][0] = 1; sum[0][1] = 0; int tot = 0; for (int i = 1; i <= n; i++) { sum[i][0] = sum[i-1][0]; sum[i][1] = sum[i-1][1]; tot += cnt[i]; if(tot&1) sum[i][1]++; else sum[i][0]++; } LL ans = 0; for (int l = 1; l <= n; l++) { int up = min(l+127, n); int mx = -0x3f3f3f3f, tot = 0; for (int i = l; i <= up; i++) { mx = max(mx, cnt[i]); tot += cnt[i]; if(tot%2 == 0 && tot >= mx*2) ans++; } } tot = 0; for (int i = 1; i <= 128; i++) tot += cnt[i]; for (int i = 129; i <= n; i++) { tot += cnt[i]; if(tot&1) ans += sum[i-129][1]; else ans += sum[i-129][0]; } printf("%lld\n", ans); return 0; }
Codeforces 1053 B - Vasya and Good Sequences
标签:and c++ force print codeforce define can main eof
原文地址:https://www.cnblogs.com/widsom/p/9737489.html