poj Problem A. Score Sequence（暴力）

  1 #include <stdio.h>
  2 #include <stdlib.h>
  3 #include <string.h>
  4 
  5 struct node
  6 {
  7     int data;
  8     int ge;
  9 };
 10 
 11 int cmp1(const void *a , const void *b)
 12 {
 13     return *(int *)a - *(int *)b;
 14 }
 15 int cmp2(const void *a , const void *b)
 16 {
 17     return *(int *)b - *(int *)a;
 18 }
 19 int cmp3( const void *a , const void *b )
 20 {
 21     struct node *c = (struct node *)a;
 22     struct node *d = (struct node *)b;
 23     if(c->ge != d->ge) return c->ge - d->ge;
 24     else return c->data - d->data;
 25 }
 26 
 27 int main()
 28 {
 29     int t, n, m;
 30 
 31 
 32     scanf("%d", &t);
 33 
 34     while(t--)
 35     {
 36 
 37         int a[35], b[35],c[35];
 38         int x[35],y[35];
 39         struct node d[35];
 40         scanf("%d%d", &n, &m);
 41         //printf("xmx\n");
 42         for(int i = 0; i < n; i++)
 43             scanf("%d", &a[i]);
 44         //printf("xmx\n");
 45         for(int i = 0; i < m; i++)
 46             scanf("%d", &b[i]);
 47 
 48         qsort(a, n, sizeof(a[0]), cmp1);
 49         qsort(b, m, sizeof(b[0]), cmp1);
 50 
 51         int  v = a[0];
 52         x[0]=a[0];
 53         int k1 = 0;
 54 
 55         for(int i = 1; i < n; i++)
 56         {
 57             if(v!=a[i])
 58             {
 59                 v = a[i];
 60                 k1++;
 61                 x[k1] = a[i];
 62             }
 63         }
 64 
 65         v = b[0];
 66         y[0]=b[0];
 67         int k2 = 0;
 68 
 69         for(int i = 1; i < m; i++)
 70         {
 71             if(v!=b[i])
 72             {
 73                 v = b[i];
 74                 k2++;
 75                 y[k2] = b[i];
 76             }
 77         }
 78 
 79 
 80         int maxn = 0, flag = 0;
 81         int i1, j1, start;
 82 
 83         for(int i = 0; i <= k1; i++)
 84         {
 85             for(int j = 0; j <= k2; j++)
 86             {
 87                 i1 = i;
 88                 j1 = j;
 89                 flag = 0;
 90                 //printf("xmx\n");
 91                 for(int k = 0; k <= k1&&i1<=k1&&j1<=k2; k++)
 92                 {
 93                     if(x[i1] == y[j1])
 94                     {
 95                         flag++;
 96                         i1++;
 97                         j1++;
 98                     }
 99                     else
100                     {
101                         break;
102                     }
103                 }
104                 if(maxn <= flag)
105                 {
106                     maxn = flag;
107                     start = i;
108                 }
109             }
110         }
111         if(maxn > 0)
112         {
113             v = x[start];
114             c[0]=x[start];
115             int k = 0;
116 
117             for(int i = 1; i < maxn; i++)
118             {
119                 if(v!=x[start+i])
120                 {
121                     v = x[start + i];
122                     k++;
123                     c[k] = x[start + i];
124                 }
125             }
126             qsort(c,k+1,sizeof(c[0]),cmp2);
127             for(int i=0; i<=k; i++)
128             {
129                 printf("%d ",c[i]);
130                 d[i].data=c[i];
131                 d[i].ge=c[i]%10;
132             }
133             printf("\n");
134             qsort(d,k+1,sizeof(d[0]),cmp3);
135             for(int i=0; i<=k; i++)
136                 printf("%d ",d[i].data);
137             printf("\n");
138 
139         }
140         else
141         {
142             printf("NONE\n");
143         }
144     }
145 
146     return 0;
147 }