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)==)

     {

          long long cnt=;

          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)==)

     {

          int sum =;

          for(int i =;i<n;i++)

          {

               int a,b;

               sf("%d%d",&a,&b);

               sum+=b-a+;

          }

          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[];

     while(sf("%s",a)==)

     {

          int len = strlen(a);

          LL cnt=;

          for(int i = ;i<len-;i++)

          {

               int sum = (a[i]-'')*+a[i+]-'';

               if(sum%==)

                    cnt+=i+;

          }

          for(int i = ;i<len;i++)

          {

               int sum = a[i]-'';

               if(sum%==)

                    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[],mp[][];

int n,k,d;

int main()

{

     while(sf("%d%d%d",&n,&k,&d)==)

     {

          mem(num,);

          mem(mp,);

          if(n>pow((double)k,(double)d))

          {

               pf("-1\n");

               continue;

          }

          for(int i =;i<n;i++)

          {

               int t = d-;

               num[t]++;

               while(num[t]==k)

               {

                    num[t]=;

                    num[t-]++;

                    t--;

               }

               memcpy(mp[i],num,sizeof(num));

          }

          for(int i = ;i<d;i++)

          {

               for(int j = ;j<n;j++)

                    pf("%d ",mp[j][i]+);

               blank;

          }

     }

}