标签：csharp nbsp zoj val getch ace 数列 return 个数
[BZOJ4103][Thu Summer Camp 2015]异或运算
试题描述
输入
第一行包含两个正整数n,m，分别表示两个数列的长度
输出
输入示例
3 3 1 2 4 7 6 5 3 1 2 1 2 2 1 2 1 3 4 2 3 2 3 4
输出示例
6 5 1
数据规模及约定
对于100%的数据，0<=Xi,Yj<2^31,
1<=u<=d<=n<=1000,
1<=l<=r<=m<=300000,
1<=k<=(d-u+1)*(r-l+1),
1<=p<=500
题解
注意到 n 很小，所以我们可以一维暴力，另一维可持久化 trie。
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <algorithm>
using namespace std;
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 maxn 1010
#define maxm 300010
#define maxnode 9600010
#define bin 30
#define oo 2147483647
int n, m, X[maxn];
int ToT, rt[maxm], ch[maxnode][2], siz[maxnode];
void update(int& y, int x, int p, int val) {
siz[y = ++ToT] = siz[x] + 1;
if(p < 0) return ;
memcpy(ch[y], ch[x], sizeof(ch[x]));
update(ch[y][val>>p&1], ch[x][val>>p&1], p - 1, val);
return ;
}
int lrt[maxn], rrt[maxn];
int qkth(int xl, int xr, int yl, int yr, int K) {
for(int i = xl; i <= xr; i++) lrt[i] = rt[yl-1], rrt[i] = rt[yr];
int ans = 0;
for(int i = bin; i >= 0; i--) {
int tmp = 0;
for(int j = xl; j <= xr; j++)
tmp += siz[ch[rrt[j]][X[j]>>i&1^1]] - siz[ch[lrt[j]][X[j]>>i&1^1]];
if(K <= tmp) {
for(int j = xl; j <= xr; j++)
lrt[j] = ch[lrt[j]][X[j]>>i&1^1],
rrt[j] = ch[rrt[j]][X[j]>>i&1^1];
ans = ans << 1 | 1;
}
else {
K -= tmp;
for(int j = xl; j <= xr; j++)
lrt[j] = ch[lrt[j]][X[j]>>i&1],
rrt[j] = ch[rrt[j]][X[j]>>i&1];
ans <<= 1;
}
}
return ans;
}
int main() {
n = read(); m = read();
for(int i = 1; i <= n; i++) X[i] = read();
for(int i = 1; i <= m; i++) update(rt[i], rt[i-1], bin, read());
int q = read();
while(q--) {
int xl = read(), xr = read(), yl = read(), yr = read(), k = read();
printf("%d\n", qkth(xl, xr, yl, yr, k));
}
return 0;
}
[BZOJ4103][Thu Summer Camp 2015]异或运算
标签：csharp nbsp zoj val getch ace 数列 return 个数
原文地址：http://www.cnblogs.com/xiao-ju-ruo-xjr/p/6741722.html
