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  • hdu 5344 MZL's xor(数学之异或)

    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≤5105,n=5105

     
    Output
    For every test.print the answer.
     
    Sample Input
    2
    3 5 5 7
    6 8 8 9
     
    Sample Output
    14 
    16
     
    Source
     

     找到了规律就好办了,发现对称性,相同的数异或后等于0,所以最后剩下自身*2来异或,注意要用long long

     1 #include<iostream>
     2 #include<cstdio>
     3 #include<cstring>
     4 #include<algorithm>
     5 #include<queue>
     6 #include<cmath>
     7 #include<stdlib.h>
     8 #include<map>
     9 using namespace std;
    10 #define ll long long 
    11 #define N 600000
    12 ll n,m,z,l;
    13 ll a[N];
    14 int main()
    15 {
    16     int t;
    17     scanf("%d",&t);
    18     while(t--){
    19         scanf("%I64d%I64d%I64d%I64d",&n,&m,&z,&l);
    20         a[1]=0;
    21         for(int i=2;i<=n;i++){
    22             a[i]=(a[i-1]*m+z)%l;
    23         }
    24         ll ans=0;
    25         for(int i=1;i<=n;i++){
    26             ans=ans^(a[i]+a[i]);
    27         }
    28         printf("%I64d
    ",ans);
    29     }
    30     return 0;
    31 }
    View Code
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  • 原文地址:https://www.cnblogs.com/UniqueColor/p/4799171.html
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