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转载请注明出处： http://www.cnblogs.com/fraud/ ——by fraud
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1517 Accepted Submission(s): 233
Special Judge
其实这题还是不容易过的，比赛的时候一看居然都过了100人了，而且发现直接贪心是不对的，于是乎乱搞了一发。。。结果居然过了。。。
这题首先只有在不能整除以及n<2m-1的时候才是NO,其余情况都是YES
然后在[2m-1，4m-2]的范围内贪心，然后之后，每2m个相互构一组
1 /** 2 * code generated by JHelper 3 * More info: https://github.com/AlexeyDmitriev/JHelper 4 * @author xyiyy @https://github.com/xyiyy 5 */ 6 7 #include <iostream> 8 #include <fstream> 9 10 //##################### 11 //Author:fraud 12 //Blog: http://www.cnblogs.com/fraud/ 13 //##################### 14 //#pragma comment(linker, "/STACK:102400000,102400000") 15 #include <iostream> 16 #include <sstream> 17 #include <ios> 18 #include <iomanip> 19 #include <functional> 20 #include <algorithm> 21 #include <vector> 22 #include <string> 23 #include <list> 24 #include <queue> 25 #include <deque> 26 #include <stack> 27 #include <set> 28 #include <map> 29 #include <cstdio> 30 #include <cstdlib> 31 #include <cmath> 32 #include <cstring> 33 #include <climits> 34 #include <cctype> 35 36 using namespace std; 37 #define pb(X) push_back(X) 38 #define rep(X, N) for(int X=0;X<N;X++) 39 #define rep2(X, L, R) for(int X=L;X<=R;X++) 40 typedef long long ll; 41 42 #define gao() out<<"NO"<<endl 43 vector<int> ans[11]; 44 int used[50]; 45 ll tot; 46 47 class hdu5355 { 48 public: 49 int ave; 50 51 void solve(std::istream &in, std::ostream &out) { 52 int n, m; 53 in >> n >> m; 54 tot = (ll) (n + 1) * n / 2; 55 if (tot % m != 0) { 56 gao(); 57 return; 58 } 59 int tmp = tot / m; 60 if (n < 2 * m - 1) { 61 gao(); 62 return; 63 } 64 int num = n / m; 65 if (tmp == n) { 66 out << "YES" << endl; 67 out << 1 << " " << n << endl; 68 rep(i, m - 1) { 69 out << 2 << " " << i + 1 << " " << n - i - 1 << endl; 70 } 71 return; 72 } 73 if (m == 1) { 74 out << "YES" << endl; 75 out << n; 76 rep2(i, 1, n)out << " " << i; 77 out << endl; 78 return; 79 } 80 rep2(i, 1, m)ans[i].clear(); 81 int f = n; 82 int c = 1; 83 /*while(n>=4*m-1){ 84 rep2(i,1,m){ 85 if(c&1) ans[i].pb(f - i + 1); 86 else ans[i].pb(f - m + i); 87 } 88 f -= m; 89 n -= m; 90 c++; 91 rep2(i,1,m){ 92 if(c&1) ans[i].pb(f - i + 1); 93 else ans[i].pb(f - m + i); 94 } 95 f -= m; 96 n -= m; 97 c++; 98 }*/ 99 c = (n + 1 - m * 2) % (m * 2) + m * 2 - 1; 100 int d = (n - c) / (2 * m); 101 for (int i = 1, j = c + 1; i <= m; i++) { 102 rep(k, d)ans[i].pb(j++), ans[i].pb(n--); 103 } 104 tot = n; 105 n = c; 106 ave = (ll) (1 + n) * n / 2 / m; 107 set<int> s; 108 rep2(i, 1, n)s.insert(i); 109 rep2(i, 1, m) { 110 rep(j, ave) { 111 auto it = s.upper_bound(ave - j); 112 ans[i].pb(*--it); 113 j += *it; 114 s.erase(it); 115 j--; 116 } 117 } 118 119 120 // clr(used,0); 121 // rep(i,tot)a[i] = n - i; 122 // if(dfs(0,0,n,1)){ 123 //if(dfs(m,n,0,0,n)){ 124 out << "YES" << endl; 125 // rep2(i,1,n)ans[used[i]].pb(i); 126 rep2(i, 1, m) { 127 int sz = ans[i].size(); 128 out << sz; 129 rep(j, sz) { 130 out << " " << ans[i][j]; 131 } 132 out << endl; 133 } 134 //}else out<<"NO"<<endl; 135 /*if(n % m != 0){ 136 gao(); 137 return; 138 } 139 if(num&1){ 140 if(m&1){ 141 out<<"YES"<<endl; 142 run(m); 143 rep2(i,1,m){ 144 out<<num; 145 rep(j,ans[i].size()){ 146 out<<" "<<ans[i][j]; 147 } 148 int last = 3 * m + i; 149 int f = 4 * m; 150 rep2(j,4,num){ 151 out<<" "<<last; 152 f += m; 153 if(j&1)last = f- m + i; 154 else last = f - i + 1; 155 } 156 out<<endl; 157 } 158 159 }else{ 160 gao(); 161 return; 162 } 163 }else{ 164 out<<"YES"<<endl; 165 rep2(i,1,m){ 166 out<<num; 167 int last = i; 168 int f = m; 169 rep(j,num){ 170 out<<" "<<last; 171 f += m; 172 if(j&1)last = f-i+1; 173 else last = f - m + i; 174 } 175 } 176 }*/ 177 178 } 179 180 bool dfs(int num, int now, int u, int m) { 181 if (now == 0) { 182 int i = tot; 183 while (used[i])i--; 184 used[i] = m; 185 if (dfs(num + 1, i, i - 1, m))return 1; 186 used[i] = 0; 187 return 0; 188 } 189 if (now == ave) { 190 if (num == tot)return 1; 191 else return dfs(num, 0, tot, m + 1); 192 } 193 for (int i = u; i > 0; i--) { 194 if (!used[i] && now + i <= ave) { 195 used[i] = m; 196 if (dfs(num + 1, now + i, i - 1, m))return 1; 197 used[i] = 0; 198 } 199 } 200 return false; 201 } 202 203 bool dfs(int m, int n, int tot, int num, int now) { 204 if (!m) { 205 return 1; 206 } 207 if (tot = ave) { 208 if (dfs(m - 1, n, 0, num, 0))return 1; 209 } 210 if (tot == 0) { 211 int i = n; 212 while (used[i])i--; 213 used[i] = m; 214 if (dfs(m, n, tot + i, num + 1, i))return 1; 215 used[i] = 0; 216 return 0; 217 } 218 for (int i = now; i > 0; i--) { 219 if (!used[i] && tot + i <= ave) { 220 used[i] = m; 221 if (dfs(m, n, tot + i, num + 1, i - 1))return 1; 222 used[i] = 0; 223 } 224 } 225 return false; 226 } 227 228 }; 229 230 int main() { 231 std::ios::sync_with_stdio(false); 232 std::cin.tie(0); 233 hdu5355 solver; 234 std::istream &in(std::cin); 235 std::ostream &out(std::cout); 236 int n; 237 in >> n; 238 for (int i = 0; i < n; ++i) { 239 solver.solve(in, out); 240 } 241 242 return 0; 243 }
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原文地址：http://www.cnblogs.com/fraud/p/4712153.html
