2016 CCPC 杭州

最新推荐文章于 2020-12-16 15:25:27 发布

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最新推荐文章于 2020-12-16 15:25:27 发布

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原文链接：http://www.cnblogs.com/foreignbill/p/7875880.html

A - ArcSoft's Office Rearrangement

均分石子。
好像怎么分答案都一样，于是模拟一遍。

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;

template<typename T> inline void read(T &x){
x=0;T f=1;char ch;do{ch=getchar();if(ch=='-')f=-1;}while(ch<'0'||ch>'9');do x=x*10+ch-'0',ch=getchar();while(ch<='9'&&ch>='0');x*=f;
}

template<typename A,typename B> inline void read(A&x,B&y){read(x);read(y);}
template<typename A,typename B,typename C> inline void read(A&x,B&y,C&z){read(x);read(y);read(z);}
template<typename A,typename B,typename C,typename D> inline void read(A&x,B&y,C&z,D&w){read(x);read(y);read(z);read(w);}

ll a[100005],n,k,x;
ll s;

int main(){
    //freopen("in.txt","r",stdin);
    int T,cas=1;
    cin>>T;
    while(T--){
        read(n,k);
        s=0;
        for(int i=1;i<=n;i++)
            read(a[i]),s+=a[i];
        if(s%k!=0){
            printf("Case #%d: -1\n",cas++);
            continue;
        }
        ll cmp = s / k;
        ll ans = 0;
        for(int i=1;i<=n;i++){
            if(a[i]==cmp) continue;
            if(a[i]<cmp){
                if(a[i]+a[i+1]<=cmp){
                    a[i+1]=a[i]+a[i+1];
                    ans++;
                } else {
                    a[i+1]=a[i+1]+a[i]-cmp;
                    ans+=2;
                }
            } else {
                ll tmp = a[i] / cmp;
                ll tt  = a[i] % cmp;
                if(tt!=0){
                    ans+=2;
                    a[i+1]+=tt;
                }
                ans+=tmp-1;
            }
        }
        printf("Case #%d: ",cas++);
        cout<<ans<<endl;
    }
    return 0;
}

B - Bomb

Tarjan之后算一下不同scc里面的cost。

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;


struct infonode {
    ll x,y,r,d;
}info[2005];


struct tarjan {
    const static int maxn = 2e4+7;
    const static int maxm = 2e6+7;
    struct Edge {
        int to,nxt;
        Edge(){}
        Edge(int _to,int _nxt):to(_to),nxt(_nxt){}
    }edge[maxm];
    int head[maxn],tot,n;
    int Low[maxn],DFN[maxn],Stack[maxn],Belong[maxn];
    int Index,top;
    int scc;
    int num[maxn];
    bool Instack[maxn];
    int Min[maxn];

    inline void init(int n) {
        tot = 0;this->n=n;
        memset(head,-1,sizeof head);
    }

    inline void addedge(int u,int v) {
        edge[tot]=Edge(v,head[u]);
        head[u]=tot++;
    }

    void Tarjan(int u) {
        int v;
        Low[u] = DFN[u] = ++Index;
        Stack[top++] = u;
        Instack[u] = true;
        for(int i=head[u];i!=-1;i=edge[i].nxt){
            v = edge[i].to;
            if(!DFN[v]){
                Tarjan(v);
                Low[u]=min(Low[u],Low[v]);
            } else if(Instack[v] && Low[u]>DFN[v]){
                Low[u]=DFN[v];
            }
        }
        if(Low[u]==DFN[u]){
            scc++;
            do{
                v=Stack[--top];
                Instack[v]=false;
                Belong[v]=scc;
                num[scc]++;
            }while(v!=u);
        }
    }

    int indeg[maxn];

    inline ll solve(int n){
        memset(DFN,0,sizeof DFN);
        memset(num,0,sizeof num);
        memset(Instack,0,sizeof Instack);
        Index = scc = top = 0;
        for(int i=1;i<=n;i++)
            if(!DFN[i]) Tarjan(i);
        memset(indeg,0,sizeof indeg);

        //if(scc==1) return 0;
        for(int u=1;u<=n;u++)
            for(int i=head[u];i!=-1;i=edge[i].nxt){
                int v=edge[i].to;
                if(Belong[u]!=Belong[v]){
                    indeg[Belong[v]]++;
                }
            }
        memset(Min,63,sizeof Min);
        for(int i=1;i<=n;i++)
            Min[Belong[i]]=min((ll)Min[Belong[i]],info[i].d);
        ll ret = 0;
        for(int i=1;i<=scc;i++)
            if(indeg[i]==0)
                ret += (ll)Min[i];
        return ret;
    }
} g;

int n,cas=1;

template<typename T> inline void read(T &x){
x=0;T f=1;char ch;do{ch=getchar();if(ch=='-')f=-1;}while(ch<'0'||ch>'9');do x=x*10+ch-'0',ch=getchar();while(ch<='9'&&ch>='0');x*=f;
}

template<typename A,typename B> inline void read(A&x,B&y){read(x);read(y);}
template<typename A,typename B,typename C> inline void read(A&x,B&y,C&z){read(x);read(y);read(z);}
template<typename A,typename B,typename C,typename D> inline void read(A&x,B&y,C&z,D&w){read(x);read(y);read(z);read(w);}

int main(){
    //freopen("in.txt","r",stdin);
    int T;
    read(T);
    while(T--){
        read(n);
        g.init(n);
        for(int i=1;i<=n;i++)
            read(info[i].x,info[i].y,info[i].r,info[i].d);
        for(int i=1;i<n;i++)
            for(int j=i+1;j<=n;j++){
                ll dist = (ll)(info[i].x-info[j].x)*(info[i].x-info[j].x)+(info[i].y-info[j].y)*(info[i].y-info[j].y);
                if(dist<=info[i].r*info[i].r){
                    g.addedge(i,j);
                }
                if(dist<=info[j].r*info[j].r){
                    g.addedge(j,i);
                }
            }
        ll ans = g.solve(n);
        printf("Case #%d: ",cas++);
        cout<<ans<<endl;
    }
    return 0;
}

C - Car

模拟分数，否则卡精度，因为涉及多次除被除除。。。

#include <bits/stdc++.h>
#define maxn 100050
using namespace std;
typedef long long LL;
int t,n;
int a[maxn];
int Case=1;
int main(){
    //freopen("in.txt","r",stdin);
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        for(int i=1;i<=n;i++) scanf("%d",&a[i]);
        LL fenzi=a[n]-a[n-1],fenmu=1;
        LL time=1;
        for(int i=n-1;i>=1;i--){
            LL d=a[i]-a[i-1];
            fenmu*=d;
            LL tmp=fenmu/fenzi;
            if(fenmu%fenzi!=0) tmp++;
            time+=tmp;
            fenzi=d;
            fenmu=tmp;
        }
        printf("Case #%d: %lld\n",Case++,time);
    }
    return 0;
}

D - Difference

每个k预处理一半，然后用中途相遇法，就是类似尺取。

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;

ll a[10][100005],b[10][100005];
ll base[11];
ll P[11][11];

template<typename T> inline void read(T &x){
x=0;T f=1;char ch;do{ch=getchar();if(ch=='-')f=-1;}while(ch<'0'||ch>'9');do x=x*10+ch-'0',ch=getchar();while(ch<='9'&&ch>='0');x*=f;
}

template<typename A,typename B> inline void read(A&x,B&y){read(x);read(y);}
template<typename A,typename B,typename C> inline void read(A&x,B&y,C&z){read(x);read(y);read(z);}
template<typename A,typename B,typename C,typename D> inline void read(A&x,B&y,C&z,D&w){read(x);read(y);read(z);read(w);}

void init(){
    base[0]=1;
    for(int i=1;i<=10;i++)
        base[i]=base[i-1]*10ll;
    for(int i=0;i<=9;i++){
        P[i][1]=i;
        for(int j=2;j<=9;j++)
            P[i][j]=P[i][j-1]*i;
    }
    for(int kk=1;kk<=9;kk++){
        for(int i=0;i<=99999;i++){
            a[kk][i]=P[i%10][kk]+P[i%100/10][kk]+P[i%1000/100][kk]+P[i%10000/1000][kk]
            +P[i%100000/10000][kk]-(ll)i*100000ll;
            b[kk][i]=P[i%10][kk]+P[i%100/10][kk]+P[i%1000/100][kk]+P[i%10000/1000][kk]
            +P[i%100000/10000][kk]-i;
        }
    }
    for(int i=1;i<=9;i++){
        sort(a[i],a[i]+100000);
        sort(b[i],b[i]+100000);
    }
}

int main(){
    //freopen("in.txt","r",stdin);
    init();
    int T,k,cas=1;
    ll x;
    cin>>T;
    while(T--){
        read(x,k);
        ll res = 0;
        for(int l=0,r=99999;l<=99999 && r;){
            if(a[k][l]+b[k][r]>x)
                r--;
            else if(a[k][l]+b[k][r]<x)
                l++;
            else {
                ll l1=0,r1=0;
                ll t1=a[k][l];
                ll t2=b[k][r];
                while(a[k][l]==t1 && l<=99999) l++,l1++;
                while(b[k][r]==t2 && r!=0) r--,r1++;
                res += l1*r1;
            }
        }
        if(x==0) res--;
        printf("Case #%d: ",cas++);
        cout<<res<<endl;
    }
    return 0;
}

F - Four Operations

贪心+枚举。

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;

int cas=1;
char s[55];

int main(){
    //freopen("in.txt","r",stdin);
    int T;
    scanf("%d",&T);
    while(T--){
        scanf("%s",s+1);
        int len=strlen(s+1);
        ll ans = -INT_MAX;
        for(int i=1;i<=len-4;i++){
            for(int j=i+1;j<=len-3;j++){
                ll A=0,B=0,C=0,D=0,E=0;
                for(int c=1;c<=i;c++)
                    A=A*10ll+s[c]-'0';
                for(int c=i+1;c<=j;c++)
                    B=B*10ll+s[c]-'0';
                C=s[j+1]-'0';
                D=s[j+2]-'0';
                for(int c=j+3;c<=len;c++)
                    E=E*10ll+s[c]-'0';
                ll tmp = A + B - C * D / E;
                ans = max(ans,tmp);
            }
        }
        printf("Case #%d: ",cas++);
        cout<<ans<<endl;
    }
    return 0;
}

K - Kingdom of Obsession

二分图匹配，建立匹配边就行。
若有交合部分，s<n
实际上可以由(1...n)-(s+1...s+n)变成(1...s)-(n+1,...,n+s)，中间部分自己匹配。

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;

const int maxn = 1005;

int matches[maxn];

int link[maxn][maxn];
bool used[maxn];
int n,s;

bool find(int x){
    for(int j=1;j<=n;j++){
        if(link[x][j] && used[j]==false){
            used[j]=1;
            if(matches[j]==0 || find(matches[j])){
                matches[j]=x;
                return 1;
            }
        }
    }
    return 0;
}

int main(){
    //freopen("in.txt","r",stdin);
    int T,cas=1;
    scanf("%d",&T);
    while(T--){
        scanf("%d%d",&n,&s);
        if(s==0){
            printf("Case #%d: Yes\n",cas++);
            continue;
        }
        if(n>s) swap(n,s);
        if(n>1000) printf("Case #%d: No\n",cas++);
        else {
            memset(link,0,sizeof link);
            memset(matches,0,sizeof matches);
            for(int i=s+1;i<=s+n;i++){
                for(int j=1;j<=n;j++)
                    if(i%j==0)
                        link[j][i-s]=1;
            }
            int res = 0;
            for(int i=1;i<=n;i++){
                memset(used,0,sizeof used);
                if(find(i)) res++;
            }
            if(res==n){
                printf("Case #%d: Yes\n",cas++);
            } else {
                printf("Case #%d: No\n",cas++);
            }
        }
    }
    return 0;
}

转载于:https://www.cnblogs.com/foreignbill/p/7875880.html