ACM-ICPC 2018 南京赛区网络预赛

轻轻松松也能拿到区域赛名额，CCPC真的好难

An Olympian Math Problem

问答

54.76%
1000ms
65536K

Alice, a student of grade

We denote

$\times 2 \times \cdots \times (k - 1) \times kk!=1×2×⋯×(k−1)×k$

We denote

$\times 1! + 2 \times 2! + \cdots +S=1×1!+2×2!+⋯+(n - 1) \times (n-1)!(n−1)×(n−1)!$

Then $____________$

You are given an integer

You have to calculate

Input

The first line contains an integer $\le 1000)T(T≤1000), denoting the number of test cases.$

For each test case, there is a line which has an integer

It is guaranteed that $\le n\le 10^{18}2≤n≤1018.$

Output

For each test case, print an integer

Hint

The first test is: $1\times 1!= 1S=1×1!=1, and 11modulo 22 is 11.$

The second test is: $1\times 1!+2 \times 2!= 5S=1×1!+2×2!=5 , and 55modulo 33 is 22.$

样例输入

2
2
3

样例输出

1
2

题目来源

ACM-ICPC 2018 南京赛区网络预赛

A题在抢首A的道路上努力奔跑，然后CE

输出n-1，因为( p -1 )! ≡ -1 ( mod p )威尔逊定理

#include<stdio.h>
#include<bits/stdc++.h>
using namespace std;
#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 ll long long
#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.);
int main()
{
    ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
    int T;
    cin>>T;
    while(T--)
    {
        ll n;
        cin>>n;
        cout<<n-1<<"\n";
    }
    return 0;
}

The writing on the wall

Feeling hungry, a cute hamster decides to order some take-away food (like fried chicken for only

However, his owner CXY thinks that take-away food is unhealthy and expensive. So she demands her hamster to fulfill a mission before ordering the take-away food. Then she brings the hamster to a wall.

The wall is covered by square ceramic tiles, which can be regarded as a

For example, the following

Such problem is quite easy for little hamster to solve, and he quickly manages to get the answer.

Seeing this, the evil girl CXY picks up a brush and paint some tiles into black, claiming that only those rectangles which don't contain any black tiles are valid and the poor hamster should only calculate the number of the valid rectangles. Now the hamster feels the problem is too difficult for him to solve, so he decides to turn to your help. Please help this little hamster solve the problem so that he can enjoy his favorite fried chicken.

Input

There are multiple test cases in the input data.

The first line contains a integer $\le 5T≤5.$

For each test case, the first line contains $\times mn×m grid, and the number of the black tiles is kk.$

For the next $x\ yx y ,denoting a black tile is on the xx-th row and yy-th column. It's guaranteed that all the positions of the black tiles are distinct.$

For all the test cases,

$\le n \le 10^5,1\le m \le 1001≤n≤105,1≤m≤100,$

$\le k \le 10^5 , 1 \le x \le n, 1 \le y \le m0≤k≤105,1≤x≤n,1≤y≤m.$

It's guaranteed that at most $\ge 20000n≥20000.$

Output

For each test case, print "Case #

Hint

The second test case looks as follows:

样例输入

样例输出

Case #1: 36
Case #2: 20

题目来源

ACM-ICPC 2018 南京赛区网络预赛

B是个51nod的原题，可以用单调栈维护，维护出这样一个玩意

然后就可以前缀和

#include<stdio.h>
#include<bits/stdc++.h>
using namespace std;
#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 ll long long
#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.);
using namespace std;
bool s[N][105];
ll a[N][105];
int b[N],sta[N],L[N],R[N],T[N];
int n, m;
int main()
{
    ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
    int cc;
    cin>>cc;
    for(int ca=1;ca<=cc;ca++)
    {
        int n,m,k;
        cin>>n>>m>>k;
        for(int i=1; i<=n; i++)for(int j=1; j<=m; j++)s[i][j]=1;
        for(int i=0,x,y; i<k; i++)cin>>x>>y,s[x][y]=0;
        memset(a,0,sizeof a),memset(b,0,sizeof b);
        b[0]=b[m+1]=-1;
        for(int i=1,tot; i<=n; i++)
        {
            tot=-1;
            sta[++tot]=0;
            for(int j=1; j<=m; j++)
            {
                if(s[i][j]==1)++b[j];
                else b[j]=0;
                while(b[sta[tot]]>b[j])--tot;
                T[j]=sta[tot];
                while(b[sta[tot]]>=b[j])--tot;
                L[j]=sta[tot],sta[++tot]=j;
            }
            tot=-1;
            sta[++tot]=m+1;
            for(int j=m; j>0; j--)
            {
                while(b[sta[tot]]>=b[j])--tot;
                R[j]=sta[tot],sta[++tot]=j;
            }
            for(int j=1,s,t; j<=m; j++)
            {
                if(b[j]&&b[T[j]]!=b[j])
                {
                    s=max(1,max(b[L[j]],b[R[j]])+1),t=R[j]-L[j]-1;
                    if(s>b[j]) continue;
                    ++a[s][t],--a[b[j]+1][t];
                }
            }
        }
        ll ans=0,tt,t;
        for(int i=1; i<=n; i++)for(int j=1; j<=m; j++)a[i][j]+=a[i-1][j];
        for(int i=1; i<=n; i++)
        {
            tt=0;
            for(int j=m; j>0; j--)t=a[i][j],a[i][j]+=a[i][j+1]+tt,tt+=t;
            for(int j=1; j<=m; j++)ans+=a[i][j];
        }
        cout<<"Case #"<<ca<<": "<<ans<<"\n";
    }
    return 0;
}

AC Challenge

Dlsj is competing in a contest with $\le 20)n(0<n≤20) problems. And he knows the answer of all of these problems.$

However, he can submit $s_isi problems, the p_{i, 1}pi,1-th, p_{i, 2}pi,2-th, ......, p_{i, s_i}pi,si-th problem before.(0 < p_{i, j} \le n,0 < j \le s_i,0 < i \le n)(0<pi,j≤n,0<j≤si,0<i≤n)After the submit of a problem, he has to wait for one minute, or cooling down time to submit another problem. As soon as the cooling down phase ended, he will submit his solution (and get "Accepted" of course) for the next problem he selected to solve or he will say that the contest is too easy and leave the arena.$

"I wonder if I can leave the contest arena when the problems are too easy for me."
"No problem."
—— CCF NOI Problem set

If he submits and passes the $\times a_i + b_it×ai+bi points. (|a_i|, |b_i| \le 10^9)(∣ai∣,∣bi∣≤109).$

Your task is to calculate the maximum number of points he can get in the contest.

Input

The first line of input contains an integer,

Then follows $s_i + 3si+3 integers, a_i,b_i,s_i,p_1,p_2,...,p_{s_i}ai,bi,si,p1,p2,...,psias described in the description above.$

Output

Output one line with one integer, the maximum number of points he can get in the contest.

Hint

In the first sample.

On the first minute, Dlsj submitted the first problem, and get $\times 5 + 6 = 111×5+6=11points.$

On the second minute, Dlsj submitted the second problem, and get $\times 4 + 5 = 132×4+5=13 points.$

On the third minute, Dlsj submitted the third problem, and get $\times 3 + 4 = 133×3+4=13points.$

On the forth minute, Dlsj submitted the forth problem, and get $\times 2 + 3 = 114×2+3=11points.$

On the fifth minute, Dlsj submitted the fifth problem, and get $\times 1 + 2 = 75×1+2=7points.$

So he can get

In the second sample, you should note that he doesn't have to solve all the problems.

样例输入1

样例输出1

样例输入2

1
-100 0 0

样例输出2

题目来源

ACM-ICPC 2018 南京赛区网络预赛

有几个属性，但是你看了n给了20就要知道要状压dp，1<<20是1024*1024

#include <bits/stdc++.h>
using namespace std;
int dp[1<<20],a[25],b[25],S[25];
int f(int x)
{
    int ans=0;
    while(x)
    {
        x-=x&-x;
        ans++;
    }
    return ans;
}
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=0,m;i<n;i++)
    {
        scanf("%d%d%d",&a[i],&b[i],&m);
        for(int j=0,x;j<m;j++)
            scanf("%d",&x),S[i]|=(1<<(x-1));
    }
    //for(int i=0;i<n;i++)
    //    printf("%d\n",S[i]);
    memset(dp,-1,sizeof dp);
    dp[0]=0;
    for(int s=0;s<1<<n;s++)
    {
        if(dp[s]==-1) continue;
        for(int i=0;i<n;i++)
            if((s&S[i])==S[i]&&(s&(1<<i))==0)
                dp[s|(1<<i)]=max(dp[s|(1<<i)],dp[s]+a[i]*(f(s)+1)+b[i]);
    }
    int ans=0;
    for(int i=0;i<1<<n;i++)
        ans=max(ans,dp[i]);
    printf("%d\n",ans);
    return 0;
}

Sum

问答

26.14%
1000ms
512000K

A square-free integer is an integer which is indivisible by any square number except $\cdot 36=2⋅3 is square-free, but 12 = 2^2 \cdot 312=22⋅3 is not, because 2^222 is a square number. Some integers could be decomposed into product of two square-free integers, there may be more than one decomposition ways. For example, 6 = 1\cdot 6=6 \cdot 1=2\cdot 3=3\cdot 2, n=ab6=1⋅6=6⋅1=2⋅3=3⋅2,n=aband n=ban=ba are considered different if a \not = ba̸=b. f(n)f(n) is the number of decomposition ways that n=abn=ab such that aa and bb are square-free integers. The problem is calculating \sum_{i = 1}^nf(i)∑i=1nf(i).$

Input

The first line contains an integer $T(T\le 20)T(T≤20), denoting the number of test cases.$

For each test case, there first line has a integer $\le 2\cdot 10^7)n(n≤2⋅107).$

Output

For each test case, print the answer $\sum_{i = 1}^n f(i)∑i=1nf(i).$

Hint

$\sum_{i = 1}^8 f(i)=f(1)+ \cdots +f(8)∑i=18f(i)=f(1)+⋯+f(8)=1+2+2+1+2+4+2+0=14=1+2+2+1+2+4+2+0=14.$

样例输入

2
5
8

样例输出

8
14

题目来源

ACM-ICPC 2018 南京赛区网络预赛

这个题还是比较难的，但是其实就是一个筛法的改变，只要一个数它出现的次数是2，就是0

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=2e7+5;
bool pri[N];
int prime[N];
ll f[N];
int tot=0;
int main()
{
    pri[1]=1,f[1]=1;
    for(int i=2; i<N; i++)
    {
        if(pri[i]==0)prime[tot++]=i,f[i]=2;
        for(int j=0,num; j<tot&&i*1LL*prime[j]<N; ++j)
        {
            num=i*prime[j],pri[num]=1;
            if(i%prime[j])f[num]=f[i]*2;
            else if(i%(1LL*prime[j]*prime[j])==0)f[num]=0;
            else
            {
                f[num]=f[num/prime[j]/prime[j]];
                break;
            }
        }
    }
    for(int i=2; i<N; i++)f[i]+=f[i-1];
    int t,n;
    scanf("%d",&t);
    while(t--)scanf("%d",&n),printf("%lld\n",f[n]);
}

但是这个可以整除分块

这个做法是sqrt(n)的

#include<stdio.h>
#include<bits/stdc++.h>
using namespace std;
#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 ll long long
#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=2e7+5,MD=1e9+7,INF=0x3f3f3f3f;
const ll LL_INF=0x3f3f3f3f3f3f3f3f;
const double eps=1e-9,e=exp(1),PI=acos(-1.);
int a[N];
int main()
{
    ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
    for(int i=2; i*i<N; i++)
        for(int j=i*i; j<N; j+=i*i)a[j]=1;
    for(int i=1; i<N; i++)a[i]+=a[i-1];
    int T;
    cin>>T;
    while(T--)
    {
        int n;
        cin>>n;
        ll ans=0;
        int l=1,r;
        while(l<=n)
        {
            r=n/(n/l)+1;
            ans+=1LL*(r-l)*(n/l)-2LL*(a[r-1]-a[l-1])*(n/l)+1LL*(a[r-1]-a[l-1])*a[n/l];
            l=r;
        }
        cout<<ans<<"\n";
    }
    return 0;
}

Magical Girl Haze

There are $v(1\le u, v\le n)v(1≤u,v≤n). Every road has a distance c_ici. Haze is a Magical Girl that lives in City 11, she can choose no more than KK roads and make their distances become 00. Now she wants to go to City NN, please help her calculate the minimum distance.$

Input

The first line has one integer $\le T\le 5)T(1≤T≤5), then following TT cases.$

For each test case, the first line has three integers

Then the following $U_i, V_i, C_iUi,Vi,Ci. There might be multiple edges between uu and vv.$

It is guaranteed that $\le 100000, M \le 200000, K \le 10N≤100000,M≤200000,K≤10,0 \le C_i \le 1e90≤Ci≤1e9. There is at least one path between City 11 and City NN.$

Output

For each test case, print the minimum distance.

样例输入

样例输出

题目来源

ACM-ICPC 2018 南京赛区网络预赛

分层最短路

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

#define ll long long
const int maxn=1e5+5,maxm=2e5+5;
ll d[11][maxn];
int n,m,k;
int head[maxn],cnt;
struct edge
{
    int v,next;
    ll w;

}edges[maxm];
struct node
{
    ll w;
    int v;
    bool operator<(const node &D)const{
        return w>D.w;
    }
};
inline bool read(int &num)
{
        char in;bool IsN=false;
        in=getchar();
        if(in==EOF) return false;
        while(in!='-'&&(in<'0'||in>'9')) in=getchar();
        if(in=='-')
        {
            IsN=true;
            num=0;
        }
        else num=in-'0';
        while(in=getchar(),in>='0'&&in<='9')
        {
                num*=10,num+=in-'0';
        }
        if(IsN) num=-num;
        return true;
}
void dij()
{
    memset(d,0x3f,sizeof d);
    priority_queue<node>q;
    q.push({0,1});
    d[0][1]=0;
    while(!q.empty())
    {
        auto u=q.top();q.pop();
        if(u.v==n)return;
        for(int i=head[u.v];i!=-1;i=edges[i].next)
        {
            edge &v=edges[i];
            for(int j=0;j<=k;j++)
            {
                if(j>=1&&d[j][v.v]>d[j-1][u.v])
                    q.push({d[j][v.v]=d[j-1][u.v],v.v});
                if(d[j][v.v]>d[j][u.v]+v.w)
                    q.push({d[j][v.v]=d[j][u.v]+v.w,v.v});
            }
        }
    }
}
int main()
{
    int t;
    read(t);
    while(t--)
    {
        memset(head,-1,sizeof head);
        cnt=0;
        read(n),read(m),read(k);
        for(int i=0,u,v,w;i<m;i++)
        {
            read(u);
            read(v);
            read(w);
            edges[cnt].v=v;
            edges[cnt].w=1LL*w;
            edges[cnt].next=head[u];
            head[u]=cnt++;
        }
        dij();
        ll minn=0x3f3f3f3f3f3f3f3f;
        for(int i=0;i<=k;i++)
            minn=min(minn,d[i][n]);
        printf("%lld\n",minn);
    }
    return 0;
}

posted @ 2018-09-06 20:22 暴力都不会的蒟蒻阅读(654) 评论(0) 编辑收藏举报

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BobHuang

ACM-ICPC 2018 南京赛区网络预赛

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BobHuang

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