Arithmetic Progressions:
/*
需要注意两个可以优化的地方：
1）预处理，找到all integers of the form p^2 + q^2
(where p and q are non-negative integers)
2）判断以a为等差数列的第一个值，b为等差数列的公差时，会不会超过(p^2 + q^2)的上界
代码中相应位置做了说明。
*/
1 #define _CRT_SECURE_NO_WARNINGS
2 #define HOME
3 #define _CRTDBG_MAP_ALLOC
4 #include <crtdbg.h>
5
6 /*
7 ID: Jming
8 PROG: ariprog
9 LANG: C++
10 */
11 #include <iostream>
12 #include <fstream>
13 #include <sstream>
14 #include <cstdlib>
15 #include <cstdio>
16 #include <cstddef>
17 #include <iterator>
18 #include <algorithm>
19 #include <string>
20 #include <locale>
21 #include <cmath>
22 #include <vector>
23 #include <cstring>
24 #include <map>
25 #include <utility>
26 #include <queue>
27 #include <stack>
28 #include <set>
29 #include <functional>
30 using namespace std;
31 typedef pair<int, int> PII;
32 typedef long long int64;
33 const int INF = 0x3f3f3f3f;
34 const int modPrime = 3046721;
35 const double eps = 1e-9;
36 const int MaxA = 250 * 250 + 250 * 250 + 10;
37 const int MaxB = 10010;
38
39 int N, M;
40
41 priority_queue<pair<int, int>, vector<pair<int, int> >, greater<pair<int, int> > > pair_ans;
42 bool isLegal[MaxA];
43
44 int limVal;
45
46
47 void ini()
48 {
49 /*
50 预处理，找到all integers of the form p^2 + q^2
51 (where p and q are non-negative integers)
52 */
53 fill(isLegal, isLegal + MaxA + 1, false);
54 int doubleVal[255];
55 for (int i = 0; i <= M; ++i)
56 {
57 doubleVal[i] = i*i;
58 }
59 for (int i = 0; i <= M; ++i)
60 {
61 for (int j = 0; j <= M; ++j)
62 {
63 isLegal[doubleVal[i] + doubleVal[j]] = true;
64 }
65 }
66
67 }
68
69
70 void Solve()
71 {
72 for (int a = 0; a <= limVal; ++a)
73 {
74 for (int b = 1; b <= limVal; ++b)
75 {
76 if (a + (N - 1)*b > limVal)
77 {
78 /*
79 判断以a为等差数列的第一个值，b为等差数列的公差时，
80 会不会超过(p^2 + q^2)的上界
81 */
82 break;
83 }
84 int nowVal = a;
85 int cnt = 0;
86 while (true)
87 {
88 if (isLegal[nowVal])
89 {
90 ++cnt;
91 }
92 else
93 {
94 break;
95 }
96 if (cnt == N)
97 {
98 pair<int, int> pII;
99 pII.first = b;
100 pII.second = a;
101 pair_ans.push(pII);
102 break;
103 }
104 nowVal += b;
105 }
106 }
107 }
108 }
109
110
111 void printAns()
112 {
113 if (pair_ans.empty())
114 {
115 puts("NONE");
116 }
117 else
118 {
119 while (!pair_ans.empty())
120 {
121 printf("%d %d\n", pair_ans.top().second, pair_ans.top().first);
122 pair_ans.pop();
123 }
124 }
125 }
126
127 int main()
128 {
129 #ifdef HOME
130 freopen("in", "r", stdin);
131 //freopen("out", "w", stdout);
132 #endif
133
134 freopen("ariprog.in", "r", stdin);
135 freopen("ariprog.out", "w", stdout);
136
137 scanf("%d %d", &N, &M);
138 limVal = M*M + M*M;
139 ini();
140 Solve();
141 printAns();
142
143
144 #ifdef HOME
145 cerr << "Time elapsed: " << clock() / CLOCKS_PER_SEC << " ms" << endl;
146 _CrtDumpMemoryLeaks();
147 #endif
148 return 0;
149 }
Mother‘s Milk:
/*
题意关键：
FJ pours milk from one bucket to another until
the second bucket is filled or the first bucket is empty.
（每一次倒牛奶都遵守这个原则）
深搜，判重
*/
1 #define _CRT_SECURE_NO_WARNINGS
2 #define HOME
3 #define _CRTDBG_MAP_ALLOC
4 #include <crtdbg.h>
5 /*
6 ID: Jming
7 PROG: milk3
8 LANG: C++
9 */
10 #include <iostream>
11 #include <fstream>
12 #include <sstream>
13 #include <cstdlib>
14 #include <cstdio>
15 #include <cstddef>
16 #include <iterator>
17 #include <algorithm>
18 #include <string>
19 #include <locale>
20 #include <cmath>
21 #include <vector>
22 #include <cstring>
23 #include <map>
24 #include <utility>
25 #include <queue>
26 #include <stack>
27 #include <set>
28 #include <functional>
29 using namespace std;
30 typedef pair<int, int> PII;
31 typedef long long int64;
32 const int INF = 0x3f3f3f3f;
33 const int modPrime = 3046721;
34 const double eps = 1e-9;
35 const int MaxN = 25;
36 const int MaxM = 10010;
37
38 int A, B, C;
39 bool isVisited[25][25][25];
40 set<int> ansSet;
41
42 void Solve(int a, int b, int c)
43 {
44 if (!isVisited[a][b][c])
45 {
46 isVisited[a][b][c] = true;
47 if (0 == a)
48 {
49 ansSet.insert(c);
50 }
51
52 // C->A
53 if (c >= (A - a))
54 {
55 Solve(A, b, c - (A - a));
56 }
57 else
58 {
59 Solve(a + c, b, 0);
60 }
61
62 // C->B
63 if (c >= (B - b))
64 {
65 Solve(a, B, c - (B - b));
66 }
67 else
68 {
69 Solve(a, b + c, 0);
70 }
71
72 // B->A
73 if (b >= (A - a))
74 {
75 Solve(A, b - (A - a), c);
76 }
77 else
78 {
79 Solve(a + b, 0, c);
80 }
81
82 // B->C
83 if (b >= (C - c))
84 {
85 Solve(a, b - (C - c), C);
86 }
87 else
88 {
89 Solve(a, 0, c + b);
90 }
91
92 // A->B
93 if (a >= (B - b))
94 {
95 Solve(a - (B - b), B, c);
96 }
97 else
98 {
99 Solve(0, b + a, c);
100 }
101
102 // A->C
103 if (a >= (C - c))
104 {
105 Solve(a - (C - c), b, C);
106 }
107 else
108 {
109 Solve(0, b, c + a);
110 }
111 }
112 }
113
114
115 int main()
116 {
117 #ifdef HOME
118 freopen("in", "r", stdin);
119 //freopen("out", "w", stdout);
120 #endif
121
122 freopen("milk3.in", "r", stdin);
123 freopen("milk3.out", "w", stdout);
124
125 scanf("%d %d %d", &A, &B, &C);
126 memset(isVisited, false, sizeof(isVisited));
127 Solve(0, 0, C);
128 for (set<int>::iterator it = ansSet.begin(); it != ansSet.end(); )
129 {
130 printf("%d", *it);
131 ++it;
132 if (it != ansSet.end())
133 {
134 printf(" ");
135 }
136 }
137 printf("\n");
138
139 #ifdef HOME
140 cerr << "Time elapsed: " << clock() / CLOCKS_PER_SEC << " ms" << endl;
141 _CrtDumpMemoryLeaks();
142 #endif
143 return 0;
144 }
时间： 2024-08-01 10:32:11