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CS Course

Time Limit: 4000/2000 MS (Java/Others)????Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1699????Accepted Submission(s): 726

Problem Description

Little A has come to college and majored in Computer and Science.

Today he has learned bit-operations in Algorithm Lessons, and he got a problem as homework.

Here is the problem:

You are giving n non-negative integers a1,a2,?,an, and some queries.

A query only contains a positive integer p, which means you
are asked to answer the result of bit-operations (and, or, xor) of all the integers except ap.

Input

There are no more than 15 test cases.

Each test case begins with two positive integers n and p
in a line, indicate the number of positive integers and the number of queries.

2≤n,q≤105

Then n non-negative integers a1,a2,?,an follows in a line, 0≤ai≤109 for each i in range[1,n].

After that there are q positive integers p1,p2,?,pqin q lines, 1≤pi≤n for each i in range[1,q].

Output

For each query p, output three non-negative integers indicates the result of bit-operations(and, or, xor) of all non-negative integers except ap in a line.

Sample Input 3 31 1 1123 Sample Output 1 1 01 1 01 1 0
Source 2017ACM/ICPC廣西邀請(qǐng)賽-重現(xiàn)賽（感謝廣西大學(xué)）
Recommend liuyiding???|???We have carefully selected several similar problems for you:??6275?6274?6273?6272?6271
?

題意：

給定n個(gè)數(shù)，然后是q次詢問(wèn)，每次詢問(wèn)為去掉第i個(gè)數(shù)，輸出剩下n-1個(gè)數(shù)與、或、異或的結(jié)果

解法1：

前綴數(shù)組和后綴數(shù)組

用p1[i]表示 [1,i] 所有數(shù)與的結(jié)果，b1[i]表示 [i,n] 所有數(shù)與的結(jié)果

#include<bits/stdc++.h> using namespace std; const int maxn=1e5+5; int a[maxn]; int p1[maxn],b1[maxn];//and int p2[maxn],b2[maxn];//or int p3[maxn],b3[maxn];//xor int main() {ios::sync_with_stdio(0);int n,q,p;while(cin>>n>>q){cin>>a[1];p1[1]=p2[1]=p3[1]=a[1];for(int i=2;i<=n;i++){cin>>a[i];p1[i]=p1[i-1]&a[i];p2[i]=p2[i-1]|a[i];p3[i]=p3[i-1]^a[i];}b1[n]=b2[n]=b3[n]=a[n];for(int i=n-1;i>=1;i--){b1[i]=b1[i+1]&a[i];b2[i]=b2[i+1]|a[i];b3[i]=b3[i+1]^a[i];}while(q--){cin>>p;if(p==1)printf("%d %d %d\n",b1[p+1],b2[p+1],b3[p+1]);else if(p==n)printf("%d %d %d\n",p1[p-1],p2[p-1],p3[p-1]);elseprintf("%d %d %d\n",p1[p-1]&b1[p+1],p2[p-1]|b2[p+1],p3[p-1]^b3[p+1]);}}return 0; }

解法2：

用num數(shù)組記錄n個(gè)數(shù)轉(zhuǎn)為二進(jìn)制后，每一位上1的個(gè)數(shù)之和，

當(dāng)去掉一個(gè)數(shù)后，更新num數(shù)組，然后遍歷num數(shù)組

如果num[i]==n-1，那么這一位與的結(jié)果為1，否則為0（與運(yùn)算：全1為1，有0為0）

如果num[i]>=1，那么這一位或的結(jié)果為1，否則為0，（或運(yùn)算：全0為0，有1為1）

對(duì)于異或運(yùn)算：

先讓n個(gè)數(shù)異或的結(jié)果為XOR，然后XOR和去掉的數(shù)字異或，結(jié)果就是n-1個(gè)數(shù)字異或的結(jié)果（異或運(yùn)算的逆運(yùn)算）

#include<bits/stdc++.h> #include<bitset> using namespace std; const int maxn=1e5+5; int a[maxn]; int num[32];int main() {ios::sync_with_stdio(0);int n,q,p;while(cin>>n>>q){int XOR=0;fill(num,num+32,0);for(int i=1;i<=n;i++){cin>>a[i];XOR^=a[i];bitset<32>bit(a[i]);for(int j=0;j<32;j++)if(bit[j]==1)num[j]++;}while(q--){cin>>p;bitset<32>bit(a[p]);bitset<32>AND;//n-1個(gè)數(shù)字與的結(jié)果bitset<32>OR;//n-1個(gè)數(shù)字或的結(jié)果for(int j=0;j<32;j++){if(num[j]-bit[j]==(n-1))//這一位有n-1個(gè)1，與的結(jié)果為1AND.set(j);//把第j位置1elseAND.reset(j);//把第j位置0if(num[j]-bit[j]>=1)//這一位上有1，或的結(jié)果為1OR.set(j);elseOR.reset(j);}printf("%d %d %d\n",AND,OR,XOR^a[p]);}}return 0; }

總結(jié)

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