SZU3

CodeForces 343A

这是第一题，像这种水题一定不要想复杂，思路不对立马换。

抓住串联和并联，可以用辗转相除法

#include <iostream>
#include <string>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <stack>
#include <queue>
#include <cctype>
#include <vector>
#include <iterator>
#include <set>
#include <map>
#include <sstream>
using namespace std;

#define mem(a,b) memset(a,b,sizeof(a))
#define pf printf
#define sf scanf
#define spf sprintf
#define pb push_back
#define debug printf("!\n")
#define INF 10000
#define MAXN 5010
#define MAX(a,b) a>b?a:b
#define blank pf("\n")
#define LL long long
#define ALL(x) x.begin(),x.end()
#define INS(x) inserter(x,x.begin())
#define pqueue priority_queue

int main()
{
     long long a,b;
     while(sf("%I64d%I64d",&a,&b)==2)
     {
          long long cnt=0;
          while(a&&b)
          {
               if(a>b)
               {
                    cnt+=a/b;
                    a%=b;
               }
               else
               {
                    cnt+=b/a;
                    b%=a;
               }
          }
          pf("%I64d\n",cnt);
     }
}

 

CodeForces - 289A

这题考看懂题意。。

#include <iostream>
#include <string>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <stack>
#include <queue>
#include <cctype>
#include <vector>
#include <iterator>
#include <set>
#include <map>
#include <sstream>
using namespace std;

#define mem(a,b) memset(a,b,sizeof(a))
#define pf printf
#define sf scanf
#define spf sprintf
#define pb push_back
#define debug printf("!\n")
#define INF 10000
#define MAXN 5010
#define MAX(a,b) a>b?a:b
#define blank pf("\n")
#define LL long long
#define ALL(x) x.begin(),x.end()
#define INS(x) inserter(x,x.begin())
#define pqueue priority_queue

int main()
{
     int n,k;
     while(sf("%d%d",&n,&k)==2)
     {
          int sum =0;
          for(int i =0;i<n;i++)
          {
               int a,b;
               sf("%d%d",&a,&b);
               sum+=b-a+1;
          }
          pf("%d\n",(k-sum%k)%k);
     }
}

 

CodeForces 628B

这也是简单题，千万别先往复杂想。只要最后两个数字能被4整除，则这个数字就能被4整除，所以枚举最后两个数字就行

#include <iostream>
#include <string>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <stack>
#include <queue>
#include <cctype>
#include <vector>
#include <iterator>
#include <set>
#include <map>
#include <sstream>
using namespace std;

#define mem(a,b) memset(a,b,sizeof(a))
#define pf printf
#define sf scanf
#define spf sprintf
#define pb push_back
#define debug printf("!\n")
#define INF 10000
#define MAXN 5010
#define MAX(a,b) a>b?a:b
#define blank pf("\n")
#define LL long long
#define ALL(x) x.begin(),x.end()
#define INS(x) inserter(x,x.begin())
#define pqueue priority_queue

int main()
{
     char a[1000000];
     while(sf("%s",a)==1)
     {
          int len = strlen(a);
          LL cnt=0;
          for(int i = 0;i<len-1;i++)
          {
               int sum = (a[i]-'0')*10+a[i+1]-'0';
               if(sum%4==0)
                    cnt+=i+1;
          }
          for(int i = 0;i<len;i++)
          {
               int sum = a[i]-'0';
               if(sum%4==0)
                    cnt++;
          }
          pf("%I64d\n",cnt);

     }
}

 

codeforces 459C

其实可以化成长度为d的k进制的全排列

#include <iostream>
#include <string>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <stack>
#include <queue>
#include <cctype>
#include <vector>
#include <iterator>
#include <set>
#include <map>
#include <sstream>
using namespace std;

#define mem(a,b) memset(a,b,sizeof(a))
#define pf printf
#define sf scanf
#define spf sprintf
#define pb push_back
#define debug printf("!\n")
#define INF 10000
#define MAXN 5010
#define MAX(a,b) a>b?a:b
#define blank pf("\n")
#define LL long long
#define ALL(x) x.begin(),x.end()
#define INS(x) inserter(x,x.begin())
#define pqueue priority_queue

int num[1005],mp[1005][1005];
int n,k,d;

int main()
{
     while(sf("%d%d%d",&n,&k,&d)==3)
     {
          mem(num,0);
          mem(mp,0);
          if(n>pow((double)k,(double)d))
          {
               pf("-1\n");
               continue;
          }
          for(int i =0;i<n;i++)
          {
               int t = d-1;
               num[t]++;
               while(num[t]==k)
               {
                    num[t]=0;
                    num[t-1]++;
                    t--;
               }
               memcpy(mp[i],num,sizeof(num));
          }
          for(int i = 0;i<d;i++)
          {
               for(int j = 0;j<n;j++)
                    pf("%d ",mp[j][i]+1);
               blank;
          }
     }
}