HDU 5344 MZL's xor（水题）

MZL's xor

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 456    Accepted Submission(s): 322

Problem Description

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

Source

2015 Multi-University Training Contest 5

点击打开链接

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<string>
#include<math.h>
#include<queue>
#include<stack>
#include<map>
#define INF 0x3f3f3f3f
#define eps 1e-6

using namespace std;

int n,m,z,l;
int v[500015];
__int64 a[500015];
int b[500015];

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d%d%d",&n,&m,&z,&l);
        memset(v,0,sizeof(v));
        v[0] = 1;
        a[0] = 1;
        int ph,pi;
        int i;
        int flag = 0;
        m = m%l;
        z = z%l;
        for(i=2;i<=n;i++)
        {
            a[i] = ((a[i-1]*m)+z)%l;
            //printf("a[%d] = %d\n",i,a[i]);
            v[a[i]]++;
        }
        int k = 1;
        for(int i=1;i<=n;i++)
        {
            if(v[a[i]]%2 == 1)
            {
                b[k] = a[i];
                v[a[i]] = 0;
                k++;
            }
        }
        __int64 sum = 0;
        __int64 sum1 = 0;
        for(int i=1;i<k;i++)
        {
            sum = b[i] + b[i];
            sum1^=sum;
        }
        printf("%I64d\n",sum1);
    }
    return 0;
}