标签:double syn code second query 字典 main ref printf
首先确定第 m 大的是谁, 建出字典树之后二分去check, 找到第 m 大之后, 在跑一次字典树去统计总和。
为什么这个要取模啊, 卡了我半天。
#include<bits/stdc++.h> #define LL long long #define LD long double #define ull unsigned 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 ALL(x) (x).begin(), (x).end() #define fio ios::sync_with_stdio(false); cin.tie(0); using namespace std; const int N = 5e4 + 10; const int inf = 0x3f3f3f3f; const LL INF = 0x3f3f3f3f3f3f3f3f; const int mod = 1e9 + 7; const double eps = 1e-8; const double PI = acos(-1); template<class T, class S> inline void add(T& a, S b) {a += b; if(a >= mod) a -= mod;} template<class T, class S> inline void sub(T& a, S b) {a -= b; if(a < 0) a += mod;} template<class T, class S> inline bool chkmax(T& a, S b) {return a < b ? a = b, true : false;} template<class T, class S> inline bool chkmin(T& a, S b) {return a > b ? a = b, true : false;} const int LOG = 30; int n, a[N]; LL m; LL sum, num; struct Trie { int ch[N * LOG][2], cnt[N * LOG]; int f[N * LOG][30]; int tot; void init() { tot = 1; } void ins(int x) { int u = 1; for(int i = LOG - 1; i >= 0; i--) { cnt[u]++; for(int j = LOG - 1; j >= 0; j--) f[u][j] += (x >> j & 1); if(!ch[u][x >> i & 1]) ch[u][x >> i & 1] = ++tot; u = ch[u][x >> i & 1]; } for(int j = LOG - 1; j >= 0; j--) f[u][j] += (x >> j & 1); cnt[u]++; } PLL query(int x, int k, int op) { int u = 1; int now = 0; LL ans = 0; LL sum = 0; for(int i = LOG - 1; i >= 0 && u; i--) { if(x >> i & 1) { if(now + (1 << i) >= k) { ans += cnt[ch[u][0]]; if(op) { for(int j = LOG - 1; j >= 0; j--) { if(x >> j & 1) sum += 1LL * (1 << j) * (cnt[ch[u][0]] - f[ch[u][0]][j]); else sum += 1LL * (1 << j) * f[ch[u][0]][j]; } } u = ch[u][1]; } else { u = ch[u][0]; now += 1 << i; } } else { if(now + (1 << i) >= k) { ans += cnt[ch[u][1]]; if(op) { for(int j = LOG - 1; j >= 0; j--) { if(x >> j & 1) sum += 1LL * (1 << j) * (cnt[ch[u][1]] - f[ch[u][1]][j]); else sum += 1LL * (1 << j) * f[ch[u][1]][j]; } } u = ch[u][0]; } else { u = ch[u][1]; now += 1 << i; } } } if(now == k) ans += cnt[u], sum += 1LL * cnt[u] * now; return mk(ans, sum); } } trie; bool check(int x, int op) { sum = 0; num = 0; for(int i = 1; i <= n; i++) { PLL tmp = trie.query(a[i], x, op); num += tmp.fi; sum += tmp.se; } num /= 2; sum /= 2; return num >= m; } int main() { trie.init(); scanf("%d%lld", &n, &m); for(int i = 1; i <= n; i++) scanf("%d", &a[i]); for(int i = 1; i <= n; i++) trie.ins(a[i]); int low = 0, high = (1 << 30) - 1, up = 0; while(low <= high) { int mid = low + high >> 1; if(check(mid, 0)) up = mid, low = mid + 1; else high = mid - 1; } check(up, 1); sum -= 1LL * (num - m) * up; printf("%lld", sum % mod); return 0; } /* */
标签:double syn code second query 字典 main ref printf
原文地址:https://www.cnblogs.com/CJLHY/p/11051151.html