20190824考试

前言:今天参加某谷的比赛,蒟蒻自闭,没远大追求回的我去搞基础,其实出题人发了题解,主要总结得失.

T1:给定一串数,分组求异或和,ans=Σ分组异或和,求ans最大.

S1:之前做过区间求异或和,开始以为是线段树,想了一下,将数二进制拆分,发现将数一个个异或就是答案.不会证明,但感性理解一下,发现分组的目就是尽可能将同位置的'1'抵消,一个个异或看似没分组,实际已将可能的同位置'1'抵消,严谨证明看题解吧.

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
using namespace std;
#define e exit(0)
#define re register
int n,ans,val[1000010];
int main()
{
    freopen("T1.in","r",stdin);
    freopen("T1.out","w",stdout);
    scanf("%d",&n);
    for(re int i=1;i<=n;++i)
        scanf("%d",&val[i]);
    for(re int i=1;i<=n;++i){
        if(i==1)
            ans=val[i];
        else ans^=val[i];
    }
    printf("%d",ans);
    return 0;
}

T2:给定实数:n,a,b,设ans = floor( pow( n , a ) + pow( n , b ) ).设有n'∈[L,R],使ans'=floor(pow( n' , a ) + pow(n' , b) ) = ans.设len = Rmax - Lmin,给定q组询问,求Σlen.

S2:二分答案拿走65pts.

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std; 
int T,seed,op;
long long st;
double n,a,b,ans,L,R,l,r;
#define e exit(0)
namespace Mker
{
//  Powered By Kawashiro_Nitori
//  Made In Gensokyo, Nihon
    #define uint unsigned int
    uint sd;int op;
    inline void init() {scanf("%u %d", &sd, &op);}
    inline uint uint_rand()
    {
        sd ^= sd << 13;
        sd ^= sd >> 7;
        sd ^= sd << 11;
        return sd;
    }
    inline double get_n()
    {
        double x = (double) (uint_rand() % 100000) / 100000;
        return x + 4;
    }
    inline double get_k()
    {
        double x = (double) (uint_rand() % 100000) / 100000;
        return (x + 1) * 5;
    }
    inline void read(double &n,double &a, double &b)
    {
        n = get_n(); a = get_k();
        if (op) b = a;
        else b = get_k(); 
    }
}
inline int fd()
{
    int s=1,t=0;
    char c=getchar();
    while(c<'0'||c>'9'){
        if(c=='-')
            s=-1;
        c=getchar();
    }
    while(c>='0'&&c<='9'){
        t=t*10+c-'0';
        c=getchar();
    }
    return s*t;
}
bool check1(double x)
{
    long long as=floor(pow(x,a)+pow(x,b));
    if(as<=st) return true;
    else if(as>st) return false;
} 
bool check2(double x)
{
    long long as=floor(pow(x,a)+pow(x,b));
    if(as>=st) return true;
    else if(as<st) return false;
}
int main()
{
    freopen("T2.in","r",stdin);
    freopen("T2.out","w",stdout);
    T=fd();
    Mker::init();
    while(T--){
        Mker::read(n,a,b);
        st=floor(pow(n,a)+pow(n,b));
        l=n-1.0,r=n+1.0;
        while(abs(r-l)>1e-10){
            double mid=(l+r)/2;
            if(check1(mid)) l=mid;
            else r=mid;
        }
        R=l;
        l=n-1.0,r=n+1.0;
        while(abs(r-l)>1e-10){
            double mid=(l+r)/2;
            if(check2(mid)) r=mid;
            else l=mid;
        }
        L=l;
        ans+=(R-L);
    }
    printf("%lf",ans);
    return 0;
}

T4:已知a0 = -3 , a1 = -6 , a2 = -12 ,an = 3an-1 + an-2 - 3an-3 + pow( 3 , n ).求ans.

S4:显然递推超时,尝试构造矩阵快速幂,无奈题目毒瘤800ms,(还以为997ms卡过了)

初始矩阵:[ a0 , a1 , a2 , 27 ].

转移矩阵:[ 0 , 0 , -3 , 0 ]

　　　　 [ 1 , 0 , 1 , 0 ]

　　　　[ 0 , 1  , 3 , 0 ]

　　　　[ 0 , 0 , 1 , 3 ]

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
#define e exit(0)
#define re register
const int mod=1e9+7;
int T;
long long n;
namespace Mker
{
//  Powered By Kawashiro_Nitori
//  Made In Gensokyo, Nihon
    #include<climits>
    #define ull unsigned long long
    #define uint unsigned int
    ull sd;int op;
    inline void init() {scanf("%llu %d", &sd, &op);}
    inline ull ull_rand()
    {
        sd ^= sd << 43;
        sd ^= sd >> 29;
        sd ^= sd << 34;
        return sd;
    }
    inline ull rand()
    {
        if (op == 0) return ull_rand() % USHRT_MAX + 1;
        if (op == 1) return ull_rand() % UINT_MAX + 1; 
        if (op == 2) return ull_rand();
    }
}
inline int fd()
{
    int x=0;int f(0);char ch(getchar());
    while(!isdigit(ch)) f|=(ch=='-'),ch=getchar();
    while(isdigit(ch)) x=(x<<1)+(x<<3)+(ch^48),ch=getchar();
    return x=f?-x:x;
}
struct jz{
    int a[10][10];
    jz(){memset(a,0,sizeof(a));}
    jz operator*(const jz p){
        jz ans;
        for(re int k=1;k<=4;++k)
            for(re int i=1;i<=4;++i)
                for(re int j=1;j<=4;++j)
                    ans.a[i][j]=((ans.a[i][j]%mod+(1ll*a[i][k]*p.a[k][j])%mod)%mod+mod)%mod;
        return ans;
    }
};
jz work()
{
    jz g;
    g.a[1][1]=0,g.a[2][1]=1,g.a[3][1]=0,g.a[4][1]=0;
    g.a[1][2]=0,g.a[2][2]=0,g.a[3][2]=1,g.a[4][2]=0;
    g.a[1][3]=-3,g.a[2][3]=1,g.a[3][3]=3,g.a[4][3]=1;
    g.a[1][4]=0,g.a[2][4]=0,g.a[3][4]=0,g.a[4][4]=3;
    return g;
}
jz qsm(jz k,int y)
{
    jz ans;
    for(re int i=1;i<=4;++i)
        ans.a[i][i]=1;
    while(y){
        if(y&1) ans=ans*k;
        k=k*k;
        y>>=1;
    }
    return ans;
}
int a0=-3,a1=-6,a2=-12,x=27,ans;
int main()
{
    freopen("T3.in","r",stdin);
    freopen("T3.out","w",stdout);
    T=fd();
    Mker::init();
    while(T--){
        n=Mker::rand();
        jz k;
        k=work();
        k=qsm(k,n-2);
        int sum=0;
        for(re int i=1;i<=4;++i){
            if(i==1)
                sum=((sum%mod+1ll*a0*k.a[1][3]%mod)%mod+mod)%mod;
            else if(i==2)
                sum=((sum%mod+1ll*a1*k.a[2][3]%mod)%mod+mod)%mod;
            else if(i==3)
                sum=((sum%mod+1ll*a2*k.a[3][3]%mod)%mod+mod)%mod;
            else if(i==4)
                sum=((sum%mod+1ll*x*k.a[4][3]%mod)%mod+mod)%mod;
        }    
        if(ans==0)
            ans=sum;
        else ans^=sum;
    }
    printf("%d",ans);
    return 0;
}

可能对当前水平能做的只有这么多,别放弃拿分!

附上:题解.