Codeforces 448(#256 (Div. 2) ) 解题报告

 1 // File Name: a.cpp
 2 // Author: darkdream
 3 // Created Time: 2014年07月17日 星期四 21时44分03秒
 4 
 5 #include<vector>
 6 #include<list>
 7 #include<map>
 8 #include<set>
 9 #include<deque>
10 #include<stack>
11 #include<bitset>
12 #include<algorithm>
13 #include<functional>
14 #include<numeric>
15 #include<utility>
16 #include<sstream>
17 #include<iostream>
18 #include<iomanip>
19 #include<cstdio>
20 #include<cmath>
21 #include<cstdlib>
22 #include<cstring>
23 #include<ctime>
24 
25 using namespace std;
26 int a[10];
27 int b[10];
28 int main(){
29     int suma = 0 ; 
30     int sumb = 0 ; 
31    for (int i = 1;i <= 3;i ++)
32    {
33        scanf("%d",&a[i]);
34      suma += a[i];
35    }
36    for(int i =1 ;i<= 3;i ++)
37    {
38      scanf("%d",&b[i]);
39      sumb += b[i] ;
40    }
41    int n; 
42    scanf("%d",&n);
43    int ta;
44    int tb;
45    if(suma % 5 == 0)
46    {
47       ta = suma/5;
48    }else {
49       ta = suma/5 + 1 ;
50    }
51    if(sumb % 10 == 0)
52    {
53       tb = sumb/10;
54    }else {
55       tb = sumb/10 + 1 ;
56    }
57   if(ta + tb <= n)
58       printf("YES\n");
59   else printf("NO\n");
60 return 0;
61 }

 1 // File Name: b.cpp
 2 // Author: darkdream
 3 // Created Time: 2014年07月17日 星期四 22时12分12秒
 4 
 5 #include<vector>
 6 #include<list>
 7 #include<map>
 8 #include<set>
 9 #include<deque>
10 #include<stack>
11 #include<bitset>
12 #include<algorithm>
13 #include<functional>
14 #include<numeric>
15 #include<utility>
16 #include<sstream>
17 #include<iostream>
18 #include<iomanip>
19 #include<cstdio>
20 #include<cmath>
21 #include<cstdlib>
22 #include<cstring>
23 #include<ctime>
24 
25 using namespace std;
26 char str[110];
27 char str1[110];
28 int dp[102][102];
29 int hs[30];
30 int hs1[30];
31 int len;
32 int len1;
33 int ok1 = 0 ;
34 int is()
35 {
36   memset(dp,0,sizeof(dp));
37   for(int i = 1;i <= len;i++)
38   {
39       for(int j = 1;j <= len1 ;j ++)
40       {
41         if(str[i-1] == str1[j-1])
42             dp[i][j] = dp[i-1][j-1] + 1;
43         else dp[i][j] = max(dp[i-1][j],dp[i][j-1]);
44       }
45   }
46  /* for(int i = 1;i <= len;i++)
47   {
48       for(int j = 1;j <= len1 ;j ++)
49       {
50         printf("%d ",dp[i][j]);
51       }
52       printf("\n");
53   }*/
54   //printf("%d\n",dp[len][len1]);
55   if(dp[len][len1] == len1)
56       return 1; 
57   return 0 ;
58 }
59 int main(){
60    scanf("%s",str);
61    scanf("%s",str1);
62    len = strlen(str);
63    len1 = strlen(str1);
64    if(len < len1)
65    {
66      printf("need tree\n");
67      return 0;
68    }
69    memset(hs,0,sizeof(hs));
70    memset(hs1,0,sizeof(hs1));
71    for(int i = 0 ;i < len;i ++)
72    {  
73         hs[str[i] -‘a‘] ++;
74    }
75    for(int i = 0 ;i < len1;i ++)
76    {  
77         hs1[str1[i] -‘a‘] ++;
78    }
79    for(int i = 0 ;i <= 26;i ++)
80    {
81       if(hs[i] < hs1[i])
82       {
83          printf("need tree\n");
84          return 0; 
85       }
86    }
87    if(is())
88    {
89       printf("automaton\n");
90    }else {
91      // printf("%d %d\n",len,len1);
92       if(len == len1)
93           printf("array\n");
94       else printf("both\n");
95    }
96 return 0;
97 }

 1 // File Name: d1.cpp
 2 // Author: darkdream
 3 // Created Time: 2014年07月18日 星期五 10时38分20秒
 4 
 5 #include<vector>
 6 #include<list>
 7 #include<map>
 8 #include<set>
 9 #include<deque>
10 #include<stack>
11 #include<bitset>
12 #include<algorithm>
13 #include<functional>
14 #include<numeric>
15 #include<utility>
16 #include<sstream>
17 #include<iostream>
18 #include<iomanip>
19 #include<cstdio>
20 #include<cmath>
21 #include<cstdlib>
22 #include<cstring>
23 #include<ctime>
24 
25 using namespace std;
26 long long n , m , k ; 
27 long long solve(long long t)
28 {
29      long long  sum = 0 ; 
30      for(int i = 1;i <= n;i ++)
31      {
32         sum += min(m,t/i);
33      }
34      return sum ; 
35 }
36 int main(){
37 
38     scanf("%lld %lld %lld",&n,&m,&k);
39     long long  l = 1,r = n*m,mid,ans;
40     while(l <= r )
41     {
42 //      printf("%lld %lld\n",l,r);
43       long long mid = (l+r) >> 1;
44       if(solve(mid) >= k )
45       {
46          r = mid - 1; 
47       }else 
48          l = mid + 1;
49     }
50     printf("%lld\n",l);
51 return 0;
52 }