[ABC261E] Many Operations(dp,位运算，打表）

[ABC261E] Many Operations – 洛谷 | 计算机科学教育新生态 (luogu.com.cn)

Problem Statement

We have a variable X and N kinds of operations that change the value of X. Operation i is represented as a pair of integers (Ti​,Ai​), and is the following operation:

• if Ti​=1, it replaces the value of X with  and X and Ai​;
• if Ti​=2, it replaces the value of X with  or X or Ai​;
• if Ti​=3, it replaces the value of X with  xor X xor Ai​.

Initialize X with the value of C and execute the following procedures in order:

• Perform Operation 1, and then print the resulting value of X.
• Next, perform Operation 1,2 in this order, and then print the value of X.
• Next, perform Operation 1,2,3 in this order, and then print the value of X.
• ⋮
• Next, perform Operation 1,2,…,N in this order, and then print the value of X.

What are and,or,xorand,or,xor?

Constraints

• 1≤N≤2×105
• 1≤Ti​≤3
• 0≤Ai​<230
• 0≤C<230
• All values in input are integers.

Input

Input is given from Standard Input in the following format:

N C
T1​ A1​
T2​ A2​
⋮
TN​ AN​

Output

Print N lines, as specified in the Problem Statement.

Sample 1

3 10
3 3
2 5
1 12

9
15
12

The initial value of X is 10.

• Operation 1 changes X to 9.
• Next, Operation 1 changes X to 10, and then Operation 2 changes it to 15.
• Next, Operation 1 changes X to 12, and then Operation 2 changes it to 13, and then Operation 3 changes it to 12.

Sample 2

9 12
1 1
2 2
3 3
1 4
2 5
3 6
1 7
2 8
3 9

0
2
1
0
5
3
3
11
2

解析 ：

用dp打表，将所有可能的状态都算出来，需要注意，这里如果不使用dp打表可能会超时（dp真的是最强的算法，可以解决的问题很多，效率还高）

集合划分：不重不漏，且需要将所有需要用到的状态体现出来

f[i][j][k] 表示：c的二进制的第 j 位是 i（0或1），且第 k 次操作所得的结果；

状态转移：

v = (a[k]>>j)&amp;1;

当 t[k]==1 : f[i][j][k]=f[i][j][k-1] &amp; v;

当 t[k]==2：f[i][j][k]=f[i][j][k-1] | v;

当 t[k]==3：f[i][j][k]=f[i][j][k-1] ^v；

#include<iostream>
#include<string>
#include<cstring>
#include<cmath>
#include<ctime>
#include<algorithm>
#include<utility>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<math.h>
#include<map>

using namespace std;
typedef long long LL;
const int N = 2e5 + 5;
int n, c, t[N], a[N], f[2][35][N];

int main() {
	cin >> n >> c;
	for (int i = 1; i <= n; i++) {
		scanf("%d%d", &amp;t[i], &amp;a[i]);
	}
	for (int i = 0; i < 2; i++) {
		for (int j = 0; j <= 30; j++) {
			f[i][j][0] = i;
			for (int k = 1; k <= n; k++) {
				int v = (a[k] >> j) &amp; 1;
				if (t[k] == 1) {
					f[i][j][k] = f[i][j][k - 1] &amp; v;
				}
				else if (t[k] == 2) {
					f[i][j][k] = f[i][j][k - 1] | v;
				}
				else
					f[i][j][k] = f[i][j][k - 1] ^ v;
			}
		}
	}
	for (int i = 1; i <= n; i++) {
		int ans = 0;
		for (int j = 0,k=c; j <= 30; j++,k>>=1) {
			ans += f[k & 1][j][i] * (1 << j);
		}
		cout << ans << endl;
		c = ans;
	}
	return 0;
}

代码007普通

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