[BZOJ3207]花神的嘲讽计划Ⅰ
试题描述
输入
输出
对于每一个嘲讽做出一个回答会尴尬输出‘Yes’,否则输出‘No’
输入示例
8 5 3 1 2 3 4 5 6 7 8 2 5 2 3 4 1 8 3 2 1 5 7 4 5 6 2 5 1 2 3 1 7 3 4 5
输出示例
No
Yes
Yes
Yes
No
数据规模及约定
题中所有数据不超过2*10^9;保证方案序列的每个数字<=N
N, M <= 105,K <= 20
题解
hash + 主席树。不知为何这题哈希不用 unsigned long long 就不行。。。
#include <iostream> #include <cstdio> #include <algorithm> #include <cmath> #include <stack> #include <vector> #include <queue> #include <cstring> #include <string> #include <map> #include <set> using namespace std; const int BufferSize = 1 << 16; char buffer[BufferSize], *Head, *Tail; inline char Getchar() { if(Head == Tail) { int l = fread(buffer, 1, BufferSize, stdin); Tail = (Head = buffer) + l; } return *Head++; } int read() { int x = 0, f = 1; char c = Getchar(); while(!isdigit(c)){ if(c == ‘-‘) f = -1; c = Getchar(); } while(isdigit(c)){ x = x * 10 + c - ‘0‘; c = Getchar(); } return x * f; } #define oo 18446744073709551615llu #define maxn 200010 #define maxnode 8000010 #define LL long long #define UL unsigned LL int ToT, sumv[maxnode], lc[maxnode], rc[maxnode]; void update(int& y, int x, UL l, UL r, UL p) { sumv[y = ++ToT] = sumv[x] + 1; if(l == r) return ; UL mid = l + (r - l >> 1); lc[y] = lc[x]; rc[y] = rc[x]; if(p <= mid) update(lc[y], lc[x], l, mid, p); else update(rc[y], rc[x], mid+1, r, p); return ; } int query(int o, UL l, UL r, UL p) { if(!o) return 0; if(l == r) return sumv[o]; UL mid = l + (r - l >> 1); if(p <= mid) return query(lc[o], l, mid, p); return query(rc[o], mid+1, r, p); } int A[maxn], rt[maxn]; UL idx[maxn], Hash[maxn]; int main() { // freopen("taunt3.in", "r", stdin); // freopen("data.out", "w", stdout); int n = read(), q = read(), k = read(); for(int i = 1; i <= n; i++) A[i] = read(); idx[0] = 1; for(int i = 1; i <= n; i++) idx[i] = idx[i-1] * 233; for(int i = 1; i <= n; i++) Hash[i] = Hash[i-1] * 233 + A[i]; for(int i = 1; i <= n - k + 1; i++) { UL tmp = Hash[i+k-1] - Hash[i-1] * idx[k]; // printf("%llu(%d)%c", tmp, ToT, i < n - k + 1 ? ‘ ‘ : ‘\n‘); update(rt[i], rt[i-1], 0, oo, tmp); } while(q--) { int ql = read(), qr = read(); UL tmp = 0; for(int i = 1; i <= k; i++) tmp = tmp * 233 + read(); if(qr - ql + 1 < k){ puts("Yes"); continue; } // printf("%d %d\n", query(rt[qr-k+1], 0, MOD - 1, tmp), query(rt[ql-1], 0, MOD - 1, tmp)); puts(query(rt[qr-k+1], 0, oo, tmp) - query(rt[ql-1], 0, oo, tmp) ? "No" : "Yes"); // */ } return 0; }
原文:http://www.cnblogs.com/xiao-ju-ruo-xjr/p/6361188.html