标签:bin amp names pll can efi ret turn using
这个容斥没想出来。。。 我好菜啊。。
f[ S ] 表示若干个数 & 的值 & S == S得 方案数, 然后用这个去容斥。
求f[ S ] 需要用SOSdp
#include<bits/stdc++.h> #define LL long long #define fi first #define se second #define mk make_pair #define PLL pair<LL, LL> #define PLI pair<LL, int> #define PII pair<int, int> #define SZ(x) ((int)x.size()) #define ull unsigned long long using namespace std; const int N = 1e6 + 7; const int inf = 0x3f3f3f3f; const LL INF = 0x3f3f3f3f3f3f3f3f; const int mod = 1e9 + 7; const double eps = 1e-8; int cnt[1<<20], bin[N], num[1<<20], n; int main() { for(int i = bin[0] = 1; i < N; i++) bin[i] = bin[i - 1] * 2 % mod; scanf("%d", &n); for(int i = 1; i <= n; i++) { int x; scanf("%d", &x); cnt[x]++; } for(int i = 0; i < 20; i++) for(int S = 0; S < (1 << 20); S++) if(S >> i & 1) cnt[S ^ (1 << i)] += cnt[S]; LL ans = bin[n]; for(int S = 1; S < (1 << 20); S++) { num[S] = num[S-(S&-S)] + 1; if(num[S] & 1) ans = (ans - bin[cnt[S]] + mod) % mod; else ans = (ans + bin[cnt[S]]) % mod; } printf("%lld\n", ans); return 0; } /* */
Codeforces Round #257 (Div. 1) D - Jzzhu and Numbers 容斥原理 + SOS dp
标签:bin amp names pll can efi ret turn using
原文地址:https://www.cnblogs.com/CJLHY/p/10356114.html