pku 2356

pku 2356

http://acm.pku.edu.cn/JudgeOnline/problem?id=2356
discuss里说用鸽巢原理，我感觉我写出来的应该是o(n)，可是程序跑了500多

 1 #include < stdio.h >
 2 #include < algorithm >
 3 using namespace std;
 4 #define Max_N  10000
 5 struct node{
 6      int  num;
 7      int  yu;
 8 };
 9 int  N;
10 int   get [Max_N];
11 struct node sum[Max_N]; 
12 bool cmp(struct node a,struct node b)
13 {
14      if (a.yu < b.yu)return  true ;
15      else   if (a.yu == b.yu)return a.num < b.num;
16      else  return  false ;
17 }
18 void solve()
19 {
20      int  i,j;
21     sum[ 0 ].yu = get [ 0 ]%N;
22     sum[ 0 ].num = 0 ;
23      for (i = 1 ;i < N;i ++ ){
24         sum[i].yu = sum[i - 1 ].yu + get [i];
25         sum[i].yu% = N;
26         sum[i].num = i;} // 第一个数是get的第一个数，第二个数是前两个数的和取余，一共是N个数，就有N个和
27      // 若其中一个和取余是0，显然成立，否则根据鸽巢原理，N个数占N - 1个位子，显然会有一样的
28     sort(sum,sum + N,cmp);
29      if (!sum[ 0 ].yu){
30         printf( " %d\n " ,sum[ 0 ].num + 1 );
31          for (i = 0 ;i <= sum[ 0 ].num;i ++ )
32             printf( " %d\n " , get [i]);
33         return;
34     }
35      else  {
36          for (i = 0 ;i < N - 1 ;i ++ )
37              if (sum[i].yu == sum[i + 1 ].yu){
38                 printf( " %d\n " ,sum[i + 1 ].num - sum[i].num);
39                  for (j = sum[i].num + 1 ;j <= sum[i + 1 ].num;j ++ )
40                     printf( " %d\n " , get [j]);
41                 return;}
42         
43     }
44     printf( " 0\n " );
45 }
46 int  main()
47 {
48      int  i;
49      while (scanf( " %d " , & N)! = EOF){
50          for (i = 0 ;i < N;i ++ ){
51             scanf( " %d " , & get [i]);
52         }
53     solve();}
54     return  0 ;
55 }

以下是经过学习别人代码后重写的代码，0ms

 1 #include < iostream >
 2 using namespace std;
 3 #define Max_N  10001
 4 int  main()
 5 {
 6      int  N,i,j;
 7      int   get [Max_N];
 8      int  sum[Max_N];
 9      int  b[Max_N];
10     sum[ 0 ] = 0 ;
11     memset(b, 0 ,sizeof(b));
12     scanf( " %d " , & N);
13      for (i = 1 ;i <= N;i ++ ){
14         scanf( " %d " , & get [i]);
15          if (!( get [i]%N)){
16             printf( " 1\n%d\n " , get [i]);break;}
17          else  {
18             sum[i] = (sum[i - 1 ] + get [i])%N;
19              if (!sum[i]){
20                 printf( " %d\n " ,i);
21                  for (j = 1 ;j <= i;j ++ )
22                     printf( " %d\n " , get [j]);
23                 break;}
24         }
25     }
26      if (i > N){
27          for (i = 1 ;i <= N;i ++ ){
28              if (!b[sum[i]])
29                 b[sum[i]] = i;
30              else {
31                 printf( " %d\n " ,i - b[sum[i]]);
32                  for (j = b[sum[i]] + 1 ;j <= i;j ++ )
33                     printf( " %d\n " , get [j]);
34                 break;}
35         }
36     }
37     return  0 ;
38 }