Kblack loves flag hdu

最新推荐文章于 2018-11-02 22:54:00 发布

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本文介绍了一个关于在n*m棋盘上随机放置k面旗帜的问题，目的是计算未放置旗帜的行和列数量。通过随机数生成算法实现旗帜位置的确定，并统计结果。

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Kblack loves flags, so he has infinite flags in his pocket.

One day, Kblack is given an

n∗m n∗m chessboard and he decides to plant flags on the chessboard where the position of each flag is described as a coordinate

(x,y) (x,y), which means that the flag is planted at the

x xth line of the

y yth row.

After planting the flags, Kblack feels sorry for those lines and rows that have no flags planted on, so he would like to know that how many lines and rows there are that have no flags planted on.

Well, Kblack, unlike you, has a date tonight, so he leaves the problem to you. please resolve the problem for him. Input

You should generate the input data in your programme.

We have a private variable

x x in the generation,which equals to

seed seed initially.When you call for a random number ranged from

[l,r] [l,r],the generation will trans

x x into

(50268147x+6082187) mod 100000007 (50268147x+6082187) mod 100000007.And then,it will return

x mod (r−l+1)+l x mod (r−l+1)+l.

The first line contains a single integer

T T refers to the number of testcases.

For each testcase,there is a single line contains 4 integers

n,m,k,seed n,m,k,seed.

Then,you need to generate the

k k flags' coordinates.

For

i=1⋯k i=1⋯k,firstly generate a random number in the range of

[1,n] [1,n].Then generate a random number in the range of

[1,m] [1,m].

You can also copy the following code and run "Init" to generate the x

(only for C++ players).

<pre>
const int _K=50268147,_B=6082187,_P=100000007;
int _X;
inline int get_rand(int _l,int _r){
_X=((long long)_K*_X+_B)%_P;
return _X%(_r-_l+1)+_l;
}
int n,m,k,seed;
int x

1000001 1000001,y

1000001 1000001;
void Init(){
  scanf("%d%d%d%d",&n,&m,&k,&seed);
  _X=seed;
  for (int i=1;i<=k;++i)
    x

i i=get_rand(1,n),
y

i i=get_rand(1,m);
}
</pre>

(1≤T≤7) (1≤T≤7),

(1≤n,m≤1000000) (1≤n,m≤1000000),

(0≤k≤1000000) (0≤k≤1000000),

(0≤seed<100000007) (0≤seed<100000007)

Output

For each testcase,print a single line contained two integers,which respectively represent the number of lines and rows that have no flags planted.

Sample Input

2
4 2 3 233
3 4 4 2333

Sample Output

2 1
1 0

Hint

the flags in the first case:$\left(4,2\right)$,$\left(1,2\right)$,$\left(1,2\right)$the flags in the second case:$\left(2,1 \right)$,$\left(2,3\right)$,$\left(3,4\right)$,$\left(3,2\right)$

题意：

kblack喜欢旗帜（flag），他的口袋里有无穷无尽的旗帜。

某天，kblack得到了一个n*m的方格棋盘，他决定把kk面旗帜插到棋盘上。

每面旗帜的位置都由一个整数对(x,y)来描述，表示该旗帜被插在了第x行第y列。

插完旗帜后，kblack突然对那些没有插过旗帜的行和列很不满，于是他想知道，有多少行、列上所有格子都没有被插过旗帜。

kblack还要把妹，于是就把这个问题丢给了你，请你帮他解决。

一道水题。。。

代码

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
const int _K=50268147,_B=6082187,_P=100000007;
int _X,x[1000005],y[1000005];
inline int get_rand(int _l,int _r){
  _X=((long long)_K*_X+_B)%_P;
  return _X%(_r-_l+1)+_l;
}
int main()
{
    int t,i;
    scanf("%d",&t);
    while(t--)
    {
        memset(x,0,sizeof(x));
        memset(y,0,sizeof(y));
        int n,m,k,seed,cnt1=0,cnt2=0,a,b;
        scanf("%d%d%d%d",&n,&m,&k,&seed);
        _X=seed;
        while(k--)
        {
            a=get_rand(1,n);
            b=get_rand(1,m);
            x[a]=1;
            y[b]=1;
            //printf("%d %d\n",a,b);
        }
        //for(i=1;i<=n;i++)
            //printf("%d ",x[i]);printf("\n");
        //for(i=1;i<=m;i++)
           // printf("%d ",y[i]);printf("\n");
        for(i=1;i<=n;i++)
        {
            if(x[i]==0)cnt1++;
        }
        for(i=1;i<=m;i++)
        {
            if(y[i]==0)cnt2++;
        }
        printf("%d %d\n",cnt1,cnt2);
    }
    return 0;
}