BZOJ2875: [Noi2012]随机数生成器

【传送门:BZOJ2875


简要题意:

  给出m,a,c,x[0],并且x数组满足x[i]=(a*x[i-1]+c)%m(i≠0)

  给出n,g,求出x[n]%g


题解:

  显然用矩乘做,不过用矩乘时,要加long long,而且要用快速乘法来处理两个数之间的乘法,不然会爆long long


参考代码:

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long LL;
struct node
{
    LL a[3][3];
    node()
    {
        memset(a,0,sizeof(a));
    }
}sum,cmp;
LL m;
LL f_mul(LL a,LL b)
{
    LL ans=0;
    while(b!=0)
    {
        if(b%2LL==1LL) ans=(ans+a)%m;
        a=(a+a)%m;b/=2LL;
    }
    return ans;
}
node chengfa(node a,node b)
{
    node c;
    for(int i=1;i<=2;i++)
    {
        for(int j=1;j<=2;j++)
        {
            for(int k=1;k<=2;k++)
            {
                c.a[i][j]=(c.a[i][j]+f_mul(a.a[i][k],b.a[k][j]))%m;
            }
        }
    }
    return c;
}
node p_mod(node a,LL b)
{
    node ans;
    ans.a[1][1]=1;ans.a[2][2]=1;
    while(b!=0)
    {
        if(b%2LL==1LL) ans=chengfa(ans,a);
        a=chengfa(a,a);
        b/=2LL;
    }
    return ans;
}
int main()
{
    LL a,c,x,n,g;
    scanf("%lld%lld%lld%lld%lld%lld",&m,&a,&c,&x,&n,&g);
    sum.a[1][1]=x;sum.a[1][2]=c;
    cmp.a[1][1]=a;
    cmp.a[2][1]=1;cmp.a[2][2]=1;
    sum=chengfa(sum,p_mod(cmp,n));
    printf("%lld\n",sum.a[1][1]%g);
    return 0;
}

 

posted @ 2018-03-25 20:57  Star_Feel  阅读(54)  评论(0编辑  收藏