[Updating] Playrix Codescapes Cup (Codeforces Round #413, rated, Div. 1 + Div. 2) 解题报告

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Problem A - Carrot Cakes

//Author: Lixiang#include<stdio.h>struct A{    int n,t,k,d;    void init(){        scanf("%d%d%d%d",&n,&t,&k,&d);    }    void work(){        int a=n/k;        if(a*k<n)a++;        a*=t;        int b=d/t;        if(b*t<d)b++;        n-=b*k;        if(n<0){            puts("NO");            return ;        }        b*=t;        int c=n/k,ans;        if(c*k<n)c++;        if(c%2==1){            c=(c+1)/2;            c*=t;            ans=d+c;        }        else{            c=c/2;            c*=t;            ans=b+c;        }        if(ans<a)puts("YES");        else puts("NO");    }}sol;int main(){    sol.init();    sol.work();    return 0;}

Problem B - T-shirt buying

模拟

//Author: Lixiang#include<stdio.h>#include<map>#include<queue>using namespace std;const int maxn=200001;struct B{    map <int,bool> used;    priority_queue <int,vector <int>,greater<int> > C[4];    int p[maxn],a[maxn],b[maxn];    int N,M;    void init(){        scanf("%d",&N);        for(int i=1;i<=N;i++)            scanf("%d",p+i);        for(int i=1;i<=N;i++)            scanf("%d",a+i);        for(int i=1;i<=N;i++)            scanf("%d",b+i);        for(int i=1;i<=N;i++){            C[a[i]].push(p[i]);            C[b[i]].push(p[i]);            used[p[i]]=1;        }    }    void work(){        scanf("%d",&M);        for(int i=1,c;i<=M;i++){            scanf("%d",&c);            while(!C[c].empty()&&!used[C[c].top()])C[c].pop();            if(C[c].empty())puts("-1");            else{                printf("%d ",C[c].top());                used[C[c].top()]=0;            }        }    }}sol;int main(){    sol.init();    sol.work();    return 0;}

Problem C - Fountains

找C+C C+D D+D三种情况中最优的
二分+贪心

//Author: Lixiang#include<stdio.h>#include<algorithm>using namespace std;const int maxn=100001;struct Fountains{    int b,p;    void init(int bb,int pp){        b=bb;p=pp;    }    bool operator<(struct Fountains b)const{        return p<b.p;    }}f[2][maxn];struct C{    int m[2][maxn];    int N,C,D,n1,n2;    void init(){        char ch;        scanf("%d%d%d",&N,&C,&D);        for(int i=1,b,p;i<=N;i++){            scanf("%d %d %c",&b,&p,&ch);            if(ch=='C')f[0][++n1].init(b,p);            else f[1][++n2].init(b,p);        }        sort(f[0]+1,f[0]+n1+1);        sort(f[1]+1,f[1]+n2+1);        for(int i=1;i<=n1;i++)            m[0][i]=max(m[0][i-1],f[0][i].b);        for(int i=1;i<=n2;i++)            m[1][i]=max(m[1][i-1],f[1][i].b);    }    int bs(int val,Fountains a[],int N){        int L=1,R=N,M,ans=0;        while(L<=R){            M=(L+R)/2;            if(a[M].p>val)R=M-1;            else{                L=M+1;                ans=M;            }        }        return ans;    }    void work(){        int i=bs(C,f[0],n1),j=bs(D,f[1],n2);        if(i==0){            int ans=0;            for(;j>=2;j--){                int id=bs(D-f[1][j].p,f[1],j-1);                if(id>0)ans=max(ans,m[1][id]+f[1][j].b);            }            printf("%d\n",ans);            return ;        }        if(j==0){            int ans=0;            for(;i>=2;i--){                int id=bs(C-f[0][i].p,f[0],i-1);                if(id>0)ans=max(ans,m[0][id]+f[0][i].b);            }            printf("%d\n",ans);            return ;        }        int ans=m[0][i]+m[1][j];        for(;i>=2;i--){            int id=bs(C-f[0][i].p,f[0],i-1);            if(id>0)ans=max(ans,m[0][id]+f[0][i].b);        }        for(;j>=2;j--){            int id=bs(D-f[1][j].p,f[1],j-1);            if(id>0)ans=max(ans,m[1][id]+f[1][j].b);        }        printf("%d\n",ans);    }}sol;int main(){    sol.init();    sol.work();    return 0;}

Problem D - field expension

D题改了10发，下次再改

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