多校

MZL's xor

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 52    Accepted Submission(s): 42

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...xorBn

 

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)modl
1≤m,z,l≤5∗105,n=5∗105

 

Output

For every test.print the answer.

 

Sample Input

23 5 5 76 8 8 9

 

Sample Output

1416

思路：当i!=j时 一定存在Ai+Aj 与Aj+Ai 两数的大小相同，相等的两个数异或为0,0余任何数异或为任何数，所以只需要考虑当i==j时即可；

code：

#include<stdio.h>const int maxx=1000010;long long a[maxx];int main(){    __int64  cas,n,m,z,l,ans,i;    scanf("%I64d",&cas);    while(cas--)    {        scanf("%lld%lld%lld%lld",&n,&m,&z,&l);        a[1]=ans=0;        for(i=2;i<=n;i++)            a[i]=(a[i-1]*m+z)%l;        for(i=2;i<=n;i++)            ans=ans^(2*a[i]);        printf("%I64d\n",ans);    }}

MZL's chemistry

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 36    Accepted Submission(s): 33

Problem Description

MZL define F(X) as the first ionization energy of the chemical element X

Now he get two chemical elements U,V,given as their atomic number,he wants to compare F(U) and F(V)

It is guaranteed that atomic numbers belongs to the given set:{1,2,3,4,..18,35,36,53,54,85,86}

It is guaranteed the two atomic numbers is either in the same period or in the same group

It is guaranteed that x≠y

 

Input

There are several test cases

For each test case,there are two numbers u,v,means the atomic numbers of the two element

 

Output

For each test case,if F(u)>F(v),print "FIRST BIGGER",else print"SECOND BIGGER"

 

Sample Input

1 25 3

 

Sample Output

SECOND BIGGERFIRST BIGGER

code：

#include<bits/stdc++.h>using namespace std;int main(){    double a[100];    a[1]=1312.0;a[2]=2372.3;    a[3]=520.2;a[4]=899.5;a[5]=800.6;a[6]=1086.6;a[7]=1402.3;a[8]=1313.9;a[9]=1681.0;a[10]=2080.7;    a[11]=495.8;a[12]=737.7;a[13]=577.5;a[14]=786.5;a[15]=1011.8;a[16]=999.6;a[17]=1250.2;a[18]=1520.6;    a[35]=1139.9;a[36]=1350.8;    a[53]=1008.4;a[54]=1170.4;    a[85]=890;a[86]=1037;    int x,y;    while(scanf("%d%d",&x,&y)!=EOF)    {        if(a[x]>a[y])            printf("FIRST BIGGER\n");        else            printf("SECOND BIGGER\n");    }    return 0;}

MZL's simple problem

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 129    Accepted Submission(s): 53

Problem Description

A simple problem
Problem Description
You have a multiple set,and now there are three kinds of operations:
1 x : add number x to set
2 : delete the minimum number (if the set is empty now,then ignore it)
3 : query the maximum number (if the set is empty now,the answer is 0)

 

Input

The first line contains a number N (N≤106),representing the number of operations.
Next N line ,each line contains one or two numbers,describe one operation.
The number in this set is not greater than 109.

 

Output

For each operation 3,output a line representing the answer.

 

Sample Input

61 21 331 31 43

 

Sample Output

34

code：

#include<bits/stdc++.h>using namespace std;int main(){    int t,sum=0,Max=-1e9-7;    scanf("%d",&t);    while(t--)    {       // cout<<Max;        int x,y;        scanf("%d",&x);        if(x==1)        {            scanf("%d",&y);            Max=max(y,Max);            sum++;        }        else if(x==2)        {            if(sum>0)                sum--;            if(sum==0)                Max=-1e9;        }        else        {            if(sum==0)                printf("0\n");            else                printf("%d\n",Max);        }    }    return 0;}