其他-HDOJ-5344-MZL's xor

MZL’s xor

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 906 Accepted Submission(s): 587

Problem Description
MZL loves xor very much.Now he gets an array A.The length of A is n.He wants to know the xor of all (Ai+Aj)(1≤i,j≤n)
The xor of an array B is defined as B1 xor B2…xor Bn

Input
Multiple test cases, the first line contains an integer T(no more than 20), indicating the number of cases.
Each test case contains four integers:n,m,z,l
A1=0,Ai=(Ai−1∗m+z) mod l
1≤m,z,l≤5∗105,n=5∗105

Output
For every test.print the answer.

Sample Input
2
3 5 5 7
6 8 8 9

Sample Output
14
16

异或的交换性和结合性。

//
// main.cpp
// 150804-1002
//
// Created by 袁子涵 on 15/8/4.
// Copyright (c) 2015年 袁子涵. All rights reserved.
//

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <stdlib.h>

using namespace std;

int T;
long long int n,m,z,l,key;
long long int num[500010];
long long int b[500010];

int main(int argc, const char * argv[]) {
    cin >> T;
    while (T--) {
        cin >> n >> m >> z >> l;
        num[0]=0;
        for (long long int i=1; i<n; i++) {
            num[i]=(num[i-1]*m+z)%l;
        }
        for(int i = 1; i <= n; i ++){
            b[i] = 2 * num[i];
            if(i == 1)
                key = b[1];
            else if(i >= 2){
                key = key ^ b[i];
            }
        }
        key ^= 0;
        cout << key << endl;
    }
    return 0;
}