ACM-ICPC 2018 徐州赛区网络预赛

徐州是真的心态崩了啊，绝望的比赛，邀请赛还是拿了一个名额的，怎么不会时光倒流啊，我真菜，赛后看看榜也挺绝望的

AHard to prepare

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After Incident, a feast is usually held in Hakurei Shrine. This time Reimu asked Kokoro to deliver a Nogaku show during the feast. To enjoy the show, every audience has to wear a Nogaku mask, and seat around as a circle.

There are N guests Reimu serves. Kokoro has $2^k2k masks numbered from 0,1,\cdots,0,1,⋯, 2^k - 12k−1, and every guest wears one of the masks. The masks have dark power of Dark Nogaku, and to prevent guests from being hurt by the power, two guests seating aside must ensure that if their masks are numbered ii and jj , then ii XNOR jj must be positive. (two guests can wear the same mask). XNOR means ~(ii^jj) and every number has kk bits. (11XNOR 1 = 11=1, 00 XNOR 0 = 10=1, 11 XNOR 0 = 00=0)$

You may have seen 《A Summer Day's dream》, a doujin Animation of Touhou Project. Things go like the anime, Suika activated her ability, and the feast will loop for infinite times. This really troubles Reimu: to not make her customers feel bored, she must prepare enough numbers of different Nogaku scenes. Reimu find that each time the same guest will seat on the same seat, and She just have to prepare a new scene for a specific mask distribution. Two distribution plans are considered different, if any guest wears different masks.

In order to save faiths for Shrine, Reimu have to calculate that to make guests not bored, how many different Nogaku scenes does Reimu and Kokoro have to prepare. Due to the number may be too large, Reimu only want to get the answer modules

Input

First line one number $\le 20)(T≤20) .$

Next $\le 1e6)k(0<N,k≤1e6) .$

Output

For each testcase output one line with a single number of scenes Reimu and Kokoro have to prepare, the answer modules

样例输入

2
3 1
4 2

样例输出

2
84

题目来源

ACM-ICPC 2018 徐州赛区网络预赛

A这个函数，你变换一下就是不让相邻的数的非异或和为0，你把问题反过来，就是相邻的异或值为2^k-1

考虑第一个数有2^k种方法，第二个数就有2^k-1种，第n个数有2^k-2种

然后第一个和最后一个还要相邻，所以呢，你可以把漏算的情况考虑一下，就是第1个数和第(n-1)个数相同，所以公式就都有了

#include<stdio.h>
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define lson l,(l+r)/2,rt<<1
#define rson (l+r)/2+1,r,rt<<1|1
#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define pb push_back
#define fi first
#define se second
#define sz(x) (int)(x).size()
#define pll pair<long long,long long>
#define pii pair<int,int>
#define pq priority_queue
const int N=1e5+5,MD=1e9+7,INF=0x3f3f3f3f;
const ll LL_INF=0x3f3f3f3f3f3f3f3f;
const double eps=1e-9,e=exp(1),PI=acos(-1.);
ll POW(ll x,ll y)
{
    ll ans=1;
    for(;y;x=x*x%MD,y>>=1)if(y&1)ans=ans*x%MD;
    return ans;
}
int main()
{
    ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
    int T;
    cin>>T;
    while(T--)
    {
        int n,k;
        cin>>n>>k;
        ll t1=POW(2,k),t2=(t1-1+MD)%MD,t2_inv=POW(t2,MD-2),t3=(t1-2+MD)%MD;
        ll sum=t1;
        for(int i=2;i<=n;i++)sum=sum*t2%MD;
        ll ans=sum;
        for(int i=2;i<=n;i++)
        {
            sum=sum*t2_inv%MD;
            if(i&1)ans=(ans-sum*t3%MD)%MD;
        }
        cout<<(ans+MD)%MD<<"\n";
    }
    return 0;
}

B. BE, GE or NE

In a world where ordinary people cannot reach, a boy named "Koutarou" and a girl named "Sena" are playing a video game. The game system of this video game is quite unique: in the process of playing this game, you need to constantly face the choice, each time you choose the game will provide

That is, if there are three options in a selection, the score will be increased by

After all the choices have been made, the score will affect the ending of the game. If the score is greater than or equal to a certain value

Koutarou and Sena are playing very happy, but I believe you have seen through the final ending. Now give you the initial score, the

Input

The first line contains four integers $n,m,k,ln,m,k,l（1\le n \le 10001≤n≤1000, -100 \le m \le 100−100≤m≤100 , -100 \le l < k \le 100−100≤l<k≤100 ）, represents the number of choices, the initial score, the minimum score required to enter a good ending, and the highest score required to enter a bad ending, respectively.$

Each of the next $a,b,ca,b,c（a\ge 0a≥0 , b\ge0b≥0 ,c=0c=0or c=1c=1）,indicates the options that appear in this selection,in which a=0a=0means there is no option to increase the score in this selection, a>0a>0 means there is an option in this selection to increase the score by aa ; b=0b=0 means there is no option to decrease the score in this selection, b>0b>0 means there is an option in this selection to decrease the score by bb; c=0c=0 means there is no option to multiply the score by -1−1 in this selection , c=1c=1 means there is exactly an option in this selection to multiply the score by -1−1. It is guaranteed that a,b,ca,b,c are not equal to 00 at the same time.$

Output

One line contains the final ending of the game. If it will enter a good ending,print "Good Ending"(without quotes); if it will enter a bad ending,print "Bad Ending"(without quotes);otherwise print "Normal Ending"(without quotes).

样例输入1

样例输出1

Good Ending

样例输入2

样例输出2

Bad Ending

题目来源

ACM-ICPC 2018 徐州赛区网络预赛

B是一个模拟题吧，你把所有情况都列出来，然后记忆化一下

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=1005;
int down,up;
int s,n;
int a[N],b[N],c[N],dp[N][305];
int R,L;
map<int,int>id;
int dfs(int pos,int now)
{
    if(pos==n+1)
    {
        if(now>=R) return 2;
        else if(now>L) return 1;
        return 0;
    }
    if(dp[pos][id[now]]!=-1) return dp[pos][id[now]];
    if(pos&1)
    {
        int f=0;
        if(a[pos])f=max(f,dfs(pos+1,min(now+a[pos],up)));
        if(b[pos])f=max(f,dfs(pos+1,max(now+b[pos],down)));
        if(c[pos])f=max(f,dfs(pos+1,-now));
        return dp[pos][id[now]]=f;
    }
    else
    {
        int f=2;
        if(a[pos])f=min(f,dfs(pos+1,min(now+a[pos],up)));
        if(b[pos])f=min(f,dfs(pos+1,max(now+b[pos],down)));
        if(c[pos])f=min(f,dfs(pos+1,-now));
        return dp[pos][id[now]]=f;
    }
}

int main()
{
    int tot=0;
    for(int i=-100; i<=100; i++) id[i]=++tot;
    down=-100,up=100;
    scanf("%d%d%d%d",&n,&s,&R,&L);
    for(int i=1; i<=n; i++)scanf("%d%d%d",&a[i],&b[i],&c[i]),b[i]=-b[i];
    memset(dp,-1,sizeof(dp));
    int f=dfs(1,s);
    if(f==2)printf("Good Ending\n");
    else if(f==1)printf("Normal Ending\n");
    else printf("Bad Ending\n");
    return 0;
}

F. Features Track

Morgana is learning computer vision, and he likes cats, too. One day he wants to find the cat movement from a cat video. To do this, he extracts cat features in each frame. A cat feature is a two-dimension vector < $x_ixi = x_jxj and y_iyi = y_jyj, then <x_ixi, y_iyi> <x_jxj, y_jyj> are same features.$

So if cat features are moving, we can think the cat is moving. If feature <

Now given the features in each frames, the number of features may be different, Morgana wants to find the longest features movement.

Input

First line contains one integer $\le T \le 10)T(1≤T≤10) , giving the test cases.$

Then the first line of each cases contains one integer

In The next $k_iki ( the number of features) and 2k_i2ki intergers describe k_iki features in ith frame.(The first two integers describe the first feature, the 33rd and 44th integer describe the second feature, and so on).$

In each test case the sum number of features $\le 100000N≤100000 .$

Output

For each cases, output one line with one integers represents the longest length of features movement.

样例输入

样例输出

题目来源

ACM-ICPC 2018 徐州赛区网络预赛

直接01滚动

#include<bits/stdc++.h>
using namespace std;
int dp[2][100005];
map<int,long long> ma[2];
int main()
{
    int T,n;
    scanf("%d",&T);
    while(T--)
    {
        memset(dp,0,sizeof dp);
        ma[0].clear(),ma[1].clear();
        scanf("%d",&n);
        int cnt=0,maxx=0;
        for(int i=1,k;i<=n;i++)
        {
            scanf("%d",&k);
            for(int j=0;j<100005;j++) dp[cnt][j]=0;
            ma[cnt].clear();
            for(int j=1,a,b;j<=k;j++)
            {
                scanf("%d%d",&a,&b);
                long long has=1LL*a*(1e9+7)+b;
                ma[cnt][has]=j;
                dp[cnt][j]=dp[cnt^1][ma[cnt^1][has]]+1;
                maxx=max(maxx,dp[cnt][j]);
            }
            cnt^=1;
        }
        printf("%d\n",maxx);
    }
    return 0;
}

G. Trace

There's a beach in the first quadrant. And from time to time, there are sea waves. A wave (

Input

The first line is the number of waves $\le 50000)n(n≤50000).$

The next $\le 10000000y≤10000000)，the ii-th line means the ii-th second there comes a wave of ( xx , yy ), it's guaranteed that when 1 \le i1≤i , j \le nj≤n ，x_i \le x_jxi≤xj and y_i \le y_jyi≤yj don't set up at the same time.$

Output

An Integer stands for the answer.

Hint：

As for the sample input, the answer is

样例输入

样例输出

题目来源

ACM-ICPC 2018 徐州赛区网络预赛

就是因为G自闭了啊，不懂怎么时光倒流

其实只差一点

#include<bits/stdc++.h>
using namespace std;
const int N=50005;
int n,m,x[N],y[N];
int main()
{
    cin>>n;
    for(int i=0; i<n; i++)scanf("%d%d",&x[i],&y[i]);
    long long ans=0;
    set<int> sx,sy;
    set<int>::iterator itx,ity;
    for(int i=n-1; i>=0; i--)
    {
        itx=sx.lower_bound(x[i]),ity=sy.lower_bound(y[i]);
        if(itx==sx.begin())ans+=x[i];
        else itx--,ans+=x[i]-*itx;
        if(ity==sy.begin())ans+=y[i];
        else ity--,ans+=y[i]-*ity;
        sx.insert(x[i]),sy.insert(y[i]);
    }
    cout<<ans<<endl;
    return 0;
}

代码可以更短

#include<bits/stdc++.h>
using namespace std;
const int N=50005;
int n,m,x[N],y[N];
int main()
{
    cin>>n;
    for(int i=0; i<n; i++)scanf("%d%d",&x[i],&y[i]);
    long long ans=0;
    set<int> sx,sy;
    sx.insert(0),sy.insert(0);
    set<int>::iterator itx,ity;
    for(int i=n-1; i>=0; i--)
    {
        itx=sx.lower_bound(x[i]),ity=sy.lower_bound(y[i]);
        itx--,ans+=x[i]-*itx;
        ity--,ans+=y[i]-*ity;
        sx.insert(x[i]),sy.insert(y[i]);
    }
    cout<<ans<<endl;
    return 0;
}

H. Ryuji doesn't want to study

Ryuji is not a good student, and he doesn't want to study. But there are n books he should learn, each book has its knowledge

Unfortunately, the longer he learns, the fewer he gets.

That means, if he reads books from $\times L + a[l+1] \times (L-1) + \cdots + a[r-1] \times 2 + a[r]a[l]×L+a[l+1]×(L−1)+⋯+a[r−1]×2+a[r](LL is the length of [ ll, rr ] that equals to r - l + 1r−l+1).$

Now Ryuji has

Input

First line contains two integers $\le 100000q≤100000).$

The next line contains n integers represent $\le 1e9)a[i](a[i]≤1e9) .$

Then in next

Output

For each question, output one line with one integer represent the answer.

样例输入

样例输出

10
8

题目来源

ACM-ICPC 2018 徐州赛区网络预赛

线段树裸题

#include<bits/stdc++.h>
using namespace std;

#define LL long long
const int maxn=1e5+5;
LL sum[maxn<<2],sum1[maxn<<2];
void pushup(int rt)
{
    sum[rt]=sum[rt<<1]+sum[rt<<1|1];
    sum1[rt]=sum1[rt<<1]+sum1[rt<<1|1];
}
void build(int l,int r,int rt)
{
    if(l==r)
    {
        scanf("%d",&sum[rt]);
        sum1[rt]=sum[rt]*l;
        return;
    }
    int mid=(l+r)>>1;
    build(l,mid,rt<<1);
    build(mid+1,r,rt<<1|1);
    pushup(rt);
}
void update(int L,LL C,int l,int r,int rt)
{
    if(l==r)
    {
        sum[rt]=C;
        sum1[rt]=sum[rt]*l;
        return;
    }
    int mid=(l+r)>>1;
    if(L<=mid)update(L,C,l,mid,rt<<1);
    else update(L,C,mid+1,r,rt<<1|1);
    pushup(rt);
}
LL query(int L,int R,int l,int r,int rt)
{
    if(L<=l&&r<=R)return sum1[rt];
    int mid=(l+r)>>1;
    LL ans=0;
    if(L<=mid)ans+=query(L,R,l,mid,rt<<1);
    if(R>mid)ans+=query(L,R,mid+1,r,rt<<1|1);
    return ans;
}
LL query1(int L,int R,int l,int r,int rt)
{
    if(L<=l&&r<=R)return sum[rt];
    int mid=(l+r)>>1;
    LL ans=0;
    if(L<=mid)ans+=query1(L,R,l,mid,rt<<1);
    if(R>mid)ans+=query1(L,R,mid+1,r,rt<<1|1);
    return ans;
}
int main()
{
    int n,q;
    scanf("%d%d",&n,&q);
    build(1,n,1);
    for(int i=0;i<q;i++)
    {
        int op,x,y;
        scanf("%d%d%d",&op,&x,&y);
        if(op==1)
        {
            printf("%lld\n",query1(x,y,1,n,1)*(y+1)-query(x,y,1,n,1));
        }
        else if(op==2)
        {
            update(x,1LL*y,1,n,1);
        }
    }
    return 0;
}

I. Characters with Hash

Mur loves hash algorithm, and he sometimes encrypt another one's name, and call him with that encrypted value. For instance, he calls Kimura KMR, and calls Suzuki YJSNPI. One day he read a book about SHA-

It's easy to find out that the algorithm cannot transfer any input string into the same length. Though in despair, Mur still wants to know the length of the answer the algorithm produces. Due to the silliness of Mur, he can even not figure out this, so you are assigned with the work to calculate the answer.

Input

First line a integer $\le 10)(T≤10).$

For each test case:

First line contains a integer $\le 1000000)(N≤1000000).$

Second line contains a string with length

Output

A single number which gives the answer.

样例输入

2
3 z
oMl
6 Y
YJSNPI

样例输出

6
10

题目来源

ACM-ICPC 2018 徐州赛区网络预赛

签到题

#include<bits/stdc++.h>
using namespace std;
const int maxn=1e6+5;
char s[maxn],has[2];
int out[maxn];
int main()
{
    int n,t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%s%s",&n,has,s);
        for(int i=0;i<n;i++)
        {
            int a=abs(has[0]-s[i]);
            out[i]=a;
        }
        int sum=n*2;
        for(int i=0;i<n;i++)
            if(out[i]==0)
                sum-=2;
            else if(out[i]<10)
            {
                sum--;
                break;
            }
            else
                break;
        if(sum==0)sum=1;
        printf("%d\n",sum);
    }
    return 0;
}

posted @ 2018-09-10 15:41 暴力都不会的蒟蒻阅读(1172) 评论(0) 编辑收藏举报

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BobHuang

ACM-ICPC 2018 徐州赛区网络预赛

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BobHuang

ACM-ICPC 2018 徐州赛区网络预赛

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