Codeforces - Gym102028 - 2018-2019 ACM-ICPC, Asia Jiaozuo Regional Contest

http://codeforces.com/gym/102028

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A. Xu Xiake in Henan Province

看起来像水题。乱搞一下，还真是。

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

int main(){
    int n;
    while(~scanf("%d",&n)){
        for(int i=0;i<n;i++){
            int cnt=0;
            for(int j=0;j<4;j++){
                int t;
                scanf("%d",&t);
                if(t)
                    cnt++;
            }

            string ans;
            switch(cnt){
            case 0:
                ans="Typically Otaku";
                break;
            case 1:
                ans="Eye-opener";
                break;
            case 2:
                ans="Young Traveller";
                break;
            case 3:
                ans="Excellent Traveller";
                break;
            case 4:
                ans="Contemporary Xu Xiake";
                break;
            }
            cout<<ans<<endl;
        }

    }
}

View Code

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I. Distance

看起来蛮暴力的？每次选离点多的那一侧最远的那个？（也就是轮流选）先莽一发。

差点被卡memset，还好发现得早……

居然忘记处理t组数据，都已经扫进来了！

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

bool c[100005];
int x[100005];

int main(){
    int t;
    while(~scanf("%d",&t)){
        while(t--){
            int n;
            scanf("%d",&n);

            memset(c,0,sizeof(c[0])*n);
            memset(x,0,sizeof(x[0])*n);

            x[0]=0;
            for(int i=0;i<n-1;i++){
                int d;
                scanf("%d",&d);
                x[i+1]=x[i]+d;
            }

            printf("%d %d",0,x[n-1]);
            c[0]=c[n-1]=1;

            int cleft=1,cright=1;

            int i=1,j=n-2;

            ll ans=x[n-1];
            ll cur=x[n-1];

            while(cleft+cright<n){
                if(cleft==cright){
                    ans+=cur;
                    if(x[i]-x[i-1]<=x[j+1]-x[j]){
                        //cur+=x[i]-x[i-1];
                        c[i]=1;
                        i++;
                        cleft++;
                    }
                    else{
                        //cur+=x[j+1]-x[j];
                        c[j]=1;
                        j--;
                        cright++;
                    }
                }
                else if(cleft<cright){
                    cur+=x[j+1]-x[i];
                    ans+=cur;

                    c[i]=1;
                    i++;
                    cleft++;
                }
                else{
                    cur+=x[j]-x[i-1];
                    ans+=cur;

                    c[j]=1;
                    j--;
                    cright++;
                }
                printf("% lld",ans);
                //printf("cur=%lld\n",cur);
            }
            printf("\n");
        }
    }
}


/*
1
8
7 6 5 4 5 6 7
*/

View Code

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E. Resistors in Parallel

感觉像是选不超过n的最小的质因数连乘形成的乘积的因数。

2,3->1/2

2,3,5->5/12=30/72

2,3,5,7->35/96=210/576