USACO chapter 1 section 1.4 clocks
USER: tianbing tianbing [tbbd4261] TASK: clocks LANG: C++ Compiling... Compile: OK Executing... Test 1: TEST OK [0.054 secs, 2932 KB] Test 2: TEST OK [0.043 secs, 2932 KB] Test 3: TEST OK [0.043 secs, 2932 KB] Test 4: TEST OK [0.054 secs, 2932 KB] Test 5: TEST OK [0.043 secs, 2932 KB] Test 6: TEST OK [0.043 secs, 2932 KB] Test 7: TEST OK [0.032 secs, 2932 KB] Test 8: TEST OK [0.032 secs, 2932 KB] Test 9: TEST OK [0.054 secs, 2932 KB] All tests OK.Your program ('clocks') produced all correct answers! This is your submission #3 for this problem. Congratulations!
Here are the test data inputs:
------- test 1 ------- 9 9 12 6 6 6 6 3 6 ------- test 2 ------- 12 9 12 9 9 9 12 9 12 ------- test 3 ------- 6 9 3 3 3 9 12 12 12 ------- test 4 ------- 9 3 9 9 9 9 9 9 9 ------- test 5 ------- 6 12 12 12 12 12 12 12 12 ------- test 6 ------- 3 12 9 6 6 6 12 12 12 ------- test 7 ------- 12 3 3 3 6 6 12 3 6 ------- test 8 ------- 12 3 9 9 12 12 3 6 9 ------- test 9 ------- 9 12 9 12 3 12 9 12 9Keep up the good work!
题意：给你九个表的初始状态，每个表只可能是3,6,9,12，然后下面有9个操作，每个操作分把几个表的指针顺时针移动90度
要求给你一个初始状态，让你给出一种方案（移动操作的组合）使移动后所有的表的指针都指向12
提交了三次，第一次compile error 没有包含string.h，下文用到了memset，低级错误；
第二次对于test无法通过。
题解：用数组a[1……9 ]分别纪录每个表移到12需要移动的次数，接受数据时a[i]=(12-a[i])/3;即可。
9重循环暴力所有情况，用cnt[1……9]记录1……9个表的移动次数，如果可以使所有表都指向12的话就保存起来。
要注意的是：每个操作不超过四次，因为四次又回到初始的状态没有意义，表移动的次数可以超过四次，比如test 3就让A表移动了6次，而a[1]==2，也就是第i个表 移动的次数cnt[i]/4==a[i]才可，这也就是我test 3 错的原因。
最后如果有多个结果的话，是输出使所有表移动次数的和最小的那个
1
/*
2 ID:tbbd4261
3 PROG:clocks
4 LANG:C++
5 */
6
7 #include < iostream >
8 #include < fstream >
9 #include < cstring >
10 #include < vector >
11 using namespace std;
12 int a[ 10 ]; // 需要移动的次数
13 int cnt[ 10 ] = { 0 };
14 void f1( int i)
15 {
16 cnt[ 1 ] += i; cnt[ 2 ] += i; cnt[ 4 ] += i; cnt[ 5 ] += i;
17 }
18
19 void f2( int i)
20 {
21 cnt[ 1 ] += i; cnt[ 2 ] += i; cnt[ 3 ] += i;
22 }
23
24 void f3( int i)
25 {
26 cnt[ 2 ] += i; cnt[ 3 ] += i; cnt[ 5 ] += i; cnt[ 6 ] += i;
27 }
28
29 void f4( int i)
30 {
31 cnt[ 1 ] += i; cnt[ 4 ] += i; cnt[ 7 ] += i;
32 }
33
34 void f5( int i)
35 {
36 cnt[ 2 ] += i; cnt[ 4 ] += i; cnt[ 5 ] += i; cnt[ 6 ] += i;cnt[ 8 ] += i;
37 }
38
39 void f6( int i)
40 {
41 cnt[ 3 ] += i; cnt[ 6 ] += i; cnt[ 9 ] += i;
42 }
43
44 void f7( int i)
45 {
46 cnt[ 4 ] += i; cnt[ 5 ] += i; cnt[ 7 ] += i; cnt[ 8 ] += i;
47 }
48
49 void f8( int i)
50 {
51 cnt[ 7 ] += i; cnt[ 8 ] += i; cnt[ 9 ] += i;
52 }
53
54 void f9( int i)
55 {
56 cnt[ 5 ] += i; cnt[ 6 ] += i; cnt[ 8 ] += i; cnt[ 9 ] += i;
57 }
58
59 int check()
60 {
61 bool is_ans = 1 ;
62 for ( int i = 1 ; i <= 9 ; i ++ )
63 {
64 if (cnt[i] % 4 != a[i])is_ans = 0 ;//原写成cnt[i]!=a[i]，这样就没有操作的组合符合条件
65 }
66 if (is_ans) return 1 ;
67 else return 0 ;
68 }
69 vector < vector < int > > vec;
70
71 void push( int i1, int i2, int i3, int i4, int i5, int i6, int i7, int i8, int i9, int & min)
72 {
73 vector < int > tem;
74 for ( int i = 1 ;i <= i1 && i1 != 0 ;i ++ )tem.push_back( 1 );
75 for ( int i = 1 ;i <= i2 && i2 != 0 ;i ++ )tem.push_back( 2 );
76 for ( int i = 1 ;i <= i3 && i3 != 0 ;i ++ )tem.push_back( 3 );
77 for ( int i = 1 ;i <= i4 && i4 != 0 ;i ++ )tem.push_back( 4 );
78 for ( int i = 1 ;i <= i5 && i5 != 0 ;i ++ )tem.push_back( 5 );
79 for ( int i = 1 ;i <= i6 && i6 != 0 ;i ++ )tem.push_back( 6 );
80 for ( int i = 1 ;i <= i7 && i7 != 0 ;i ++ )tem.push_back( 7 );
81 for ( int i = 1 ;i <= i8 && i8 != 0 ;i ++ )tem.push_back( 8 );
82 for ( int i = 1 ;i <= i9 && i9 != 0 ;i ++ )tem.push_back( 9 );
83 vec.push_back(tem);
84 // int sum=tem.size();
85 // if(sum<min){ min=sum; vec.clear(); vec.push_back(tem); }
86 // else if(sum==min)vec.push_back(tem);
87 }
88
89 int main()
90 {
91 ifstream cin( " clocks.in " );
92 ofstream cout( " clocks.out " );
93 int min = 10000 ;
94 for ( int i = 1 ; i <= 9 ; i ++ )
95 { cin >> a[i]; a[i] = ( 12 - a[i]) / 3 ; }
96 for ( int i1 = 0 ; i1 < 4 ; i1 ++ )
97 for ( int i2 = 0 ; i2 < 4 ; i2 ++ )
98 for ( int i3 = 0 ; i3 < 4 ; i3 ++ )
99 for ( int i4 = 0 ; i4 < 4 ; i4 ++ )
100 for ( int i5 = 0 ; i5 < 4 ; i5 ++ )
101 for ( int i6 = 0 ; i6 < 4 ; i6 ++ )
102 for ( int i7 = 0 ; i7 < 4 ; i7 ++ )
103 for ( int i8 = 0 ; i8 < 4 ; i8 ++ )
104 for ( int i9 = 0 ; i9 < 4 ; i9 ++ )
105 {
106 f1(i1); f2(i2); f3(i3); f4(i4); f5(i5); f6(i6); f7(i7); f8(i8); f9(i9);
107 int v = check();
108 if (v == 1 )
109 push(i1,i2,i3,i4,i5,i6,i7,i8,i9,min);
110 memset(cnt, 0 , sizeof cnt);
111 }
112
113 int mov[ 10 ] = { 0 , 4 , 3 , 4 , 3 , 5 , 3 , 4 , 3 , 4 };
114 vector < int > temp; int m = 10000 ;
115
116 for ( int i = 0 ; i < vec.size(); i ++ )
117 {
118 int sum = 0 ;
119 for ( int j = 0 ; j < vec[i].size();j ++ )
120 sum += mov[vec[i][j]];
121 if (sum < m)
122 { m = sum; temp = vec[i]; }
123 }
124 cout << temp[ 0 ];
125 for ( int i = 1 ; i < temp.size(); i ++ )
126 cout << ' ' << temp[i];
127 cout << endl;
128
129 // system("pause");
130 return 0 ;
131 }
132
2 ID:tbbd4261
3 PROG:clocks
4 LANG:C++
5 */
6
7 #include < iostream >
8 #include < fstream >
9 #include < cstring >
10 #include < vector >
11 using namespace std;
12 int a[ 10 ]; // 需要移动的次数
13 int cnt[ 10 ] = { 0 };
14 void f1( int i)
15 {
16 cnt[ 1 ] += i; cnt[ 2 ] += i; cnt[ 4 ] += i; cnt[ 5 ] += i;
17 }
18
19 void f2( int i)
20 {
21 cnt[ 1 ] += i; cnt[ 2 ] += i; cnt[ 3 ] += i;
22 }
23
24 void f3( int i)
25 {
26 cnt[ 2 ] += i; cnt[ 3 ] += i; cnt[ 5 ] += i; cnt[ 6 ] += i;
27 }
28
29 void f4( int i)
30 {
31 cnt[ 1 ] += i; cnt[ 4 ] += i; cnt[ 7 ] += i;
32 }
33
34 void f5( int i)
35 {
36 cnt[ 2 ] += i; cnt[ 4 ] += i; cnt[ 5 ] += i; cnt[ 6 ] += i;cnt[ 8 ] += i;
37 }
38
39 void f6( int i)
40 {
41 cnt[ 3 ] += i; cnt[ 6 ] += i; cnt[ 9 ] += i;
42 }
43
44 void f7( int i)
45 {
46 cnt[ 4 ] += i; cnt[ 5 ] += i; cnt[ 7 ] += i; cnt[ 8 ] += i;
47 }
48
49 void f8( int i)
50 {
51 cnt[ 7 ] += i; cnt[ 8 ] += i; cnt[ 9 ] += i;
52 }
53
54 void f9( int i)
55 {
56 cnt[ 5 ] += i; cnt[ 6 ] += i; cnt[ 8 ] += i; cnt[ 9 ] += i;
57 }
58
59 int check()
60 {
61 bool is_ans = 1 ;
62 for ( int i = 1 ; i <= 9 ; i ++ )
63 {
64 if (cnt[i] % 4 != a[i])is_ans = 0 ;//原写成cnt[i]!=a[i]，这样就没有操作的组合符合条件
65 }
66 if (is_ans) return 1 ;
67 else return 0 ;
68 }
69 vector < vector < int > > vec;
70
71 void push( int i1, int i2, int i3, int i4, int i5, int i6, int i7, int i8, int i9, int & min)
72 {
73 vector < int > tem;
74 for ( int i = 1 ;i <= i1 && i1 != 0 ;i ++ )tem.push_back( 1 );
75 for ( int i = 1 ;i <= i2 && i2 != 0 ;i ++ )tem.push_back( 2 );
76 for ( int i = 1 ;i <= i3 && i3 != 0 ;i ++ )tem.push_back( 3 );
77 for ( int i = 1 ;i <= i4 && i4 != 0 ;i ++ )tem.push_back( 4 );
78 for ( int i = 1 ;i <= i5 && i5 != 0 ;i ++ )tem.push_back( 5 );
79 for ( int i = 1 ;i <= i6 && i6 != 0 ;i ++ )tem.push_back( 6 );
80 for ( int i = 1 ;i <= i7 && i7 != 0 ;i ++ )tem.push_back( 7 );
81 for ( int i = 1 ;i <= i8 && i8 != 0 ;i ++ )tem.push_back( 8 );
82 for ( int i = 1 ;i <= i9 && i9 != 0 ;i ++ )tem.push_back( 9 );
83 vec.push_back(tem);
84 // int sum=tem.size();
85 // if(sum<min){ min=sum; vec.clear(); vec.push_back(tem); }
86 // else if(sum==min)vec.push_back(tem);
87 }
88
89 int main()
90 {
91 ifstream cin( " clocks.in " );
92 ofstream cout( " clocks.out " );
93 int min = 10000 ;
94 for ( int i = 1 ; i <= 9 ; i ++ )
95 { cin >> a[i]; a[i] = ( 12 - a[i]) / 3 ; }
96 for ( int i1 = 0 ; i1 < 4 ; i1 ++ )
97 for ( int i2 = 0 ; i2 < 4 ; i2 ++ )
98 for ( int i3 = 0 ; i3 < 4 ; i3 ++ )
99 for ( int i4 = 0 ; i4 < 4 ; i4 ++ )
100 for ( int i5 = 0 ; i5 < 4 ; i5 ++ )
101 for ( int i6 = 0 ; i6 < 4 ; i6 ++ )
102 for ( int i7 = 0 ; i7 < 4 ; i7 ++ )
103 for ( int i8 = 0 ; i8 < 4 ; i8 ++ )
104 for ( int i9 = 0 ; i9 < 4 ; i9 ++ )
105 {
106 f1(i1); f2(i2); f3(i3); f4(i4); f5(i5); f6(i6); f7(i7); f8(i8); f9(i9);
107 int v = check();
108 if (v == 1 )
109 push(i1,i2,i3,i4,i5,i6,i7,i8,i9,min);
110 memset(cnt, 0 , sizeof cnt);
111 }
112
113 int mov[ 10 ] = { 0 , 4 , 3 , 4 , 3 , 5 , 3 , 4 , 3 , 4 };
114 vector < int > temp; int m = 10000 ;
115
116 for ( int i = 0 ; i < vec.size(); i ++ )
117 {
118 int sum = 0 ;
119 for ( int j = 0 ; j < vec[i].size();j ++ )
120 sum += mov[vec[i][j]];
121 if (sum < m)
122 { m = sum; temp = vec[i]; }
123 }
124 cout << temp[ 0 ];
125 for ( int i = 1 ; i < temp.size(); i ++ )
126 cout << ' ' << temp[i];
127 cout << endl;
128
129 // system("pause");
130 return 0 ;
131 }
132