codeforces #298

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今天小练了一套题~这一套题好像都是思维题

A Exam

n个学生排成一排编号1,2,3....n，那么会发现第i个学生编号与第i+1个学生编号相邻，这样不满足要求。题目要求任意相邻的两学生编号差值的绝对值不为1；

思路：

把编号为奇数的抽出来为一组 1,3,5,7,9....

把编号为偶数的抽出来为一组 2,4,6,8,10...

然后两者都反向排列即可 ....9 7 5 3 1 .....10 8 6 4 2 注意判一下偶数组的最大值和奇数组的最小值的大小关系

#include <iostream>#include <stdio.h>#include <string>#include <string.h>#include <algorithm>using namespace std;int a[2550],b[2550];int ct1,ct2;int abs(int a){  return (a>0)?a:(-a);}int main(){    int n;    while(cin>>n)    {       ct1=ct2=0;       for( int i=1; i<=n; i++)       {          if(i&1)          {               a[ct1++]=i;          }          else b[ct2++]=i;       }       if( abs(b[ ct2-1 ]-a[0])==1 )       {            printf("%d\n",ct1);            for(int i=ct1-1;i>=0;i--)                if(i==ct1-1)printf("%d",a[i]);                else printf(" %d",a[i]);            printf("\n");       }       else       {           printf("%d\n",ct1+ct2);            for(int i=ct1-1;i>=0;i--)                if(i==ct1-1)printf("%d",a[i]);                else printf(" %d",a[i]);            for(int i=ct2-1;i>=0;i--)                printf(" %d",b[i]);            printf("\n");       }    }    return 0;}

B题 DP搞就行，dp[i][j]表示第i秒速度为j时的最大值，第i秒由第i-1秒推过来

#include <iostream>#include <cstdio>#include <string>#include <string.h>#include <algorithm>using namespace std;int dp[110][5500];int main(){   int v1,v2,t,d;   while(cin>>v1>>v2>>t>>d)   {       memset(dp,0,sizeof(dp));       dp[1][v1]=1*v1;       for(int i=1;i<t;i++)       {           for(int v=1;v<=1200;v++)           {              if(dp[i][v]==0)continue;              for(int j=-d;j<=d;j++)              {                 if(v+j<=0)continue;                 dp[i+1][v+j]=max(dp[i+1][v+j],dp[i][v]+v+j);              }           }       }       printf("%d\n",dp[t][v2]);   }   return 0;}

C题思维题稍微计算一下每个骰子取值范围就行

#include <iostream>#include <cstdio>#include <string>#include <string.h>#include <algorithm>using namespace std;typedef long long ll;const int maxn=1e6+100;ll d[maxn];ll ans[maxn];int main(){    int n;    ll s;    while(scanf("%d%I64d",&n,&s)!=EOF)    {        ll sum=0;        for(int i=1;i<=n;i++)        {            scanf("%I64d",&d[i]);            sum+=d[i];        }        memset(ans,0,sizeof(ans));        for(int i=1;i<=n;i++)        {           ll tmp=sum-d[i];           ll left=s-tmp;           ll right=s-n+1;           if( left>1&&left<=d[i] )ans[i]+=left-1;           if(right<=0)ans[i]+=d[i];           else if(right<d[i])ans[i]+=d[i]-right;        }       for(int i=1;i<=n;i++)        if(i==1)printf("%I64d",ans[i]);        else printf(" %I64d",ans[i]);      printf("\n");    }    return 0;}

.D题贪心处理即可用vector比较方便些。

#include <iostream>#include <cstdio>#include <string>#include <string.h>#include <algorithm>#include <vector>using namespace std;const int maxn=1e5+100;vector<int>vec[2*maxn];vector<int>ans;int main(){     int n;     while(scanf("%d",&n)!=EOF)     {       for(int i=0;i<2*maxn-10;i++)vec[i].clear();       ans.clear();       for(int i=1;i<=n;i++)       {          int tmp;          scanf("%d",&tmp);          vec[tmp].push_back(i);       }       int sum=0,flag=0;       if( vec[0].empty() )       {          printf("Impossible\n");          continue;       }       ans.push_back(vec[0].back());       vec[0].pop_back();       sum++;       flag++;       for( int i=2;i<=n;i++ )       {             if(!vec[sum].empty())             {                ans.push_back(vec[sum].back());                vec[sum].pop_back();                flag++;                sum++;             }             else if(sum>=3)             {                 while(sum>=3)                 {                   sum-=3;                   if(!vec[sum].empty())                   {                    ans.push_back(vec[sum].back());                    vec[sum].pop_back();                    flag++;                    sum++;                    break;                   }                 }             }             else                break;       }       if(flag!=n)       {          printf("Impossible\n");       }       else       {          printf("Possible\n");          for(int i=0;i<ans.size();i++)            if(i==0)printf("%d",ans[i]);            else printf(" %d",ans[i]);          printf("\n");       }     }     return 0;}

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