模拟3

 

 

		ID	Origin	Title
	6 / 12	Problem A	ZOJ 3860	Find the Spy
	6 / 27	Problem B	ZOJ 3861	Valid Pattern Lock
		Problem C	ZOJ 3862	Intersection
	0 / 1	Problem D	ZOJ 3863	Paths on the Tree
		Problem E	ZOJ 3864	Quiz for EXO-L
		Problem F	ZOJ 3865	Superbot
	5 / 15	Problem G	ZOJ 3866	Cylinder Candy
	7 / 15	Problem H	ZOJ 3867	Earthstone: Easy Version
	1 / 6	Problem I	ZOJ 3868	GCD Expectation

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

A.

d.找出不同的数。。

#include<iostream>
#include<stdio.h>
using namespace std;

int main(){

    int T;
    int N;
    int ID[128];
    int i;
    int sum1,sum2;
    int p1,p2;

    scanf("%d",&T);

    while(T--){
        scanf("%d",&N);

        for(i=0;i<N;++i){
            scanf("%d",&ID[i]);
        }

        p1=0;
        for(i=1;i<N;++i){
            if(ID[i]!=ID[i-1]){
                p2=i;
                break;
            }
        }

        sum1=0;
        sum2=0;
        for(i=0;i<N;++i){
            if(ID[i]==ID[p1]){
                ++sum1;
            }
            else if(ID[i]==ID[p2]){
                ++sum2;
            }
        }

        if(sum1==1){
            printf("%d
",ID[p1]);
        }
        else{
            printf("%d
",ID[p2]);
        }
    }

    return 0;
}

B.

d.手机滑屏解锁的问题，给出几个1~9之间的几个数，求出密码的所有可能。

s.dfs

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;

bool exist[16];
bool vis[16];
int can[16][16];

int t_path[212345][10];

int path[16];

int n;

int sum;

void dfs(int u,int now){

    int i;

    if(now>n){

        for(i=1;i<=9;++i){
            t_path[sum][i]=path[i];
        }

        ++sum;
        return;
    }


    for(i=1;i<=9;++i){
        if(exist[i]&&!vis[i]){

            if( (u==1&&(i==3||i==9||i==7)) || (u==3&&(i==1||i==7||i==9)) ||
                (u==9&&(i==7||i==1||i==3)) || (u==7&&(i==9||i==3||i==1)) ){

                if(vis[can[u][i]]){
                    path[now]=i;
                    vis[i]=true;
                    dfs(i,now+1);
                    vis[i]=false;
                }
                else{

                }
            }
            else if( (u==2&&i==8) || (u==8&&i==2) ||
                     (u==4&&i==6) || (u==6&&i==4) ){

                if(vis[can[u][i]]){
                    path[now]=i;
                    vis[i]=true;
                    dfs(i,now+1);
                    vis[i]=false;
                }
                else{

                }
            }
            else{
                path[now]=i;
                vis[i]=true;
                dfs(i,now+1);
                vis[i]=false;
            }

        }
    }

}

void init(){
    memset(exist,false,sizeof(exist));
}

int main(){

    memset(can,0,sizeof(can));
    can[1][3]=2;
    can[1][9]=5;
    can[1][7]=4;

    can[3][1]=2;
    can[3][7]=5;
    can[3][9]=6;

    can[9][7]=8;
    can[9][1]=5;
    can[9][3]=6;

    can[7][9]=8;
    can[7][3]=5;
    can[7][1]=4;

    //---
    can[2][8]=5;
    can[8][2]=5;

    can[4][6]=5;
    can[6][4]=5;


    int T;
    int i;
    int t;
    int j;

    scanf("%d",&T);

    while(T--){
        scanf("%d",&n);

        init();
        for(i=0;i<n;++i){
            scanf("%d",&t);
            exist[t]=true;
        }

        sum=0;
        for(i=1;i<=9;++i){
            memset(vis,false,sizeof(vis));
            if(exist[i]){
                path[1]=i;
                vis[i]=true;
                dfs(i,2);
                vis[i]=false;
            }
        }


        printf("%d
",sum);

        for(i=0;i<sum;++i){
            for(j=1;j<n;++j){
                printf("%d ",t_path[i][j]);
            }
            printf("%d
",t_path[i][n]);
        }

    }

    return 0;
}

C.

d.线段相交拆分问题

s.先排序，好像还可以。。

 

D.

 

E.

d.黑白图像的问题

ps:好像用连通块来判断，还有黑白点的比例。题目比较长，没读题。

 

F.

 

G.

d.积分求体积，求面积。

#include <iostream>
#include <cmath>
#include <cstdio>
using namespace std;
#define PI M_PI
int main()
{
    int T;
    double r,h,d,v,s;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%lf%lf%lf",&r,&h,&d);
        v=h*PI*(r+d)*(r+d)+PI*r*r*d*2+2*2*PI*(1.0/3*d*d*d+d*d/4*PI*r);
        s=2*(PI*r*r+PI*(r+d)*h+2*PI*d*d+PI*PI*r*d);
        printf("%.12f %.12f
",v,s);
    }
    return 0;
}

H.

d.炉石传说也来水题了。。

#include<iostream>
#include<stdio.h>
using namespace std;

int main(){

    int T;
    int A1,H1;
    int A2,H2;
    int H3,H4;

    scanf("%d",&T);

    while(T--){
        scanf("%d%d%d%d",&A1,&H1,&A2,&H2);

        if(A1==0){
            printf("Invalid
");
            continue;
        }

        H3=H1-A2;
        H4=H2-A1;

        if(H3<=0){
            printf("Discard ");
        }
        else{
            printf("%d %d ",A1,H3);
        }

        if(H4<=0){
            printf("Discard
");
        }
        else{
            printf("%d %d
",A2,H4);
        }

    }

    return 0;
}

I.

ps:又是最大公约数的问题，尴尬。。

d.给出n和k, 求出n个数的任意非空子集的最大公约数的k次方的期望, 最后求出期望*(2^n-1)

s.容斥原理

每取一个子集的概率都为 1/(2^n-1), 结果除以(2^n-1),则实际上是求出每个非空子集的k次方的和.

题目转化为gcd为i的子集为多少, 枚举i 从MAX到1.

对于每个i, 求出n个数中为i 的倍数为cnt个, 

则gcd=i 的非空子集的个数dp[i] = (2^cnt-1) - dp[j] (j为i 的倍数).

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>

using namespace std;
typedef long long ll;
const int maxn = 1e6+5;
const ll mod = 998244353;
int sum[maxn];
ll dp[maxn];

ll cal(ll x, int n)
{
    ll s = 1;
    while(n > 0) {
        if(n & 1) s = (s*x) % mod;
        x = (x*x)%mod;
        n >>= 1;
    }
    return s;
}
int main()
{
//freopen("in", "r", stdin);
    int T;
    scanf("%d", &T);
    while(T--) {
        int n, k;
        scanf("%d %d", &n, &k);
        memset(sum, 0, sizeof(sum));
        memset(dp, 0, sizeof(dp));

        int MAX = 0;
        for(int i = 0; i < n; ++i) {
            int a;
            scanf("%d", &a);
            MAX = max(MAX, a);
            sum[a]++;
        }
        ll ans = 0;
        for(int i = MAX; i >= 1; --i) {
            int cnt = 0;
            for(int j = i; j <= MAX; j+=i) {
                cnt += sum[j];
                dp[i] = (dp[i] - dp[j] + mod) % mod;
            }
            dp[i] = ((dp[i]+cal(2, cnt)-1)%mod + mod) % mod;
            ans = (ans + dp[i]*cal(i, k)) % mod;
        }
        printf("%lld
", ans);

    }
    return 0;
}

c2.上个莫比乌斯反演的代码，还不懂

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <set>
#include <queue>
#include <vector>
#define LL long long
#define MOD 998244353

using namespace std;

const int maxN = 1e6+10;
int n, k, len;
int pow2[maxN], num[maxN];
int prime[maxN], u[maxN];
bool vis[maxN];

//莫比乌斯反演
//F(n)和f(n)是定义在非负整数集合上的两个函数
//并且满足条件F(n) = sum(f(d)) (d|n)
//那么f(n) = sum(u(d)*F(n/d)) (d|n)
//case 1: if d = 1, u(d) = 1
//case 2: if d = p1*p2...pk, (pi均为互异素数), u(d) = (-1)^k
//case 3: else, u(d) = 0;
//性质1: sum(u(d)) (d|n) = 1 (n == 1) or 0 (n > 1)
//性质2: sum(u(d)/d) (d|n) = phi(n)/n
//另一种形式:F(d) = sum(f(n)) (d|n) => f(d) = sum(u(n/d)*F(n)) (d|n)
//线性筛选求莫比乌斯反演函数代码
void mobius()
{
    memset(vis, false,sizeof(vis));
    u[1] = 1;
    int cnt = 0;
    for(int i = 2; i < maxN; i++)
    {
        if(!vis[i])
        {
            prime[cnt++] = i;
            u[i] = -1;
        }
        for(int j = 0; j < cnt && i*prime[j] < maxN; j++)
        {
            vis[i*prime[j]] = true;
            if(i%prime[j])
                u[i*prime[j]] = -u[i];
            else
            {
                u[i*prime[j]] = 0;
                break;
            }
        }
    }
}

void init()
{
    mobius();
    pow2[0] = 1;
    for (int i = 1; i < maxN; ++i)
        pow2[i] = 2*pow2[i-1]%MOD;
}

void input()
{
    int tmp;
    len = 0;
    scanf("%d%d", &n, &k);
    memset(num, 0, sizeof(num));
    for (int i = 0; i < n; ++i)
    {
        scanf("%d", &tmp);
        len = max(len, tmp);
        num[tmp]++;
    }
    for (int i = 1; i <= len; ++i)
    {
        for (int j = i*2; j <= len; j += i)
            num[i] += num[j];
    }
}

//快速幂m^n
LL quickPow(LL x, int n)
{
    if (n == 0)
        return 1;
    if (x == 0)
        return 0;
    LL a = 1;
    while (n)
    {
        a *= n&1 ? x : 1;
        a %= MOD;
        n >>= 1 ;
        x *= x;
        x %= MOD;
    }
    return a;
}

void work()
{
    int ans = 0, val;
    for (int i = 1; i <= len; ++i)
    {
        val = quickPow(i, k);
        for (int j = i; j <= len; j += i)
        {
            ans += ((LL)u[j/i]*val%MOD)*(pow2[num[j]]-1)%MOD;
            ans = (ans%MOD+MOD)%MOD;
        }
    }
    printf("%d
", ans);
}

int main()
{
    //freopen("test.in", "r", stdin);
    init();
    int T;
    scanf("%d", &T);
    for (int times = 1; times <= T; ++times)
    {
        input();
        work();
    }
    return 0;
}