codeforces #298

标签：codeforce #298

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;
}

#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;
}

#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;
}

#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;
}

codeforces #298

标签：codeforce #298

原文地址：http://blog.csdn.net/liusuangeng/article/details/45341835