There is the following puzzle popular among nuclear physicists.
A reactor contains a set of n atoms of some chemical elements. We shall understand the phrase "atomic number" as the number of this atom's element in the periodic table of the chemical elements.
You are allowed to take any two different atoms and fuse a new one from them. That results in a new atom, whose number is equal to the sum of the numbers of original atoms. The fusion operation can be performed several times.
The aim is getting a new pregiven set of k atoms.
The puzzle's difficulty is that it is only allowed to fuse two atoms into one, it is not allowed to split an atom into several atoms. You are suggested to try to solve the puzzle.
Output
If it is impossible to synthesize the required atoms, print "NO" without the quotes. Otherwise, print on the first line «YES», and on the next k lines print the way of synthesizing each of k atoms as equations. Each equation has the following form: "x1+x2+...+xt->yi", where xj is the symbol of the element of some atom from the original set, and yi is the symbol of the element of some atom from the resulting set. Each atom from the input data should occur in the output data exactly one time. The order of summands in the equations, as well as the output order does not matter. If there are several solutions, print any of them. For a better understanding of the output format, see the samples.
Sample test(s)
Output
YES
Mn+C+K->Sn
Co+Zn+Sc->Pt
Li+Mg+P+F->Y
Note
The reactions from the first example possess the following form (the atomic number is written below and to the left of the element):
To find a periodic table of the chemical elements, you may use your favorite search engine.
The pretest set contains each of the first 100 elements of the periodic table at least once. You can use that information to check for misprints.
學(xué)習(xí)了
fura2 的代碼——本來只是想偷懶拷貝一下元素表的，一不小心看到了代碼，于是。。。
因為學(xué)習(xí)了代碼，感覺思路還是挺簡單的，動態(tài)規(guī)劃。。
1 #include <iostream>
2 #include <cstring>
3 #include <string>
4 #include <map>
5
6 using namespace std;
7
8 const int N = 20;
9
10 int main() {
11 int n, n2, n21, k, i, j, s, t, nt;
12 static int sum[ 1<<N ], f[ 1<<N ], p[ 1<<N ];
13 string nuclearA[ N ], nuclearB[ N ];
14 int numberA[ N ], numberB[ N ];
15
16 map< string, int > number;
17 string nuclear[] = {
18 "H","He","Li","Be","B","C","N","O","F","Ne","Na","Mg","Al","Si","P","S","Cl","Ar",
19 "K","Ca","Sc","Ti","V","Cr","Mn","Fe","Co","Ni","Cu","Zn","Ga","Ge","As","Se","Br",
20 "Kr","Rb","Sr","Y","Zr","Nb","Mo","Tc","Ru","Rh","Pd","Ag","Cd","In","Sn","Sb","Te",
21 "I","Xe","Cs","Ba","La","Ce","Pr","Nd","Pm","Sm","Eu","Gd","Tb","Dy","Ho","Er","Tm",
22 "Yb","Lu","Hf","Ta","W","Re","Os","Ir","Pt","Au","Hg","Tl","Pb","Bi","Po","At","Rn",
23 "Fr","Ra","Ac","Th","Pa","U","Np","Pu","Am","Cm","Bk","Cf","Es","Fm"
24 };
25 for ( i = 0; i < sizeof(nuclear)/sizeof(nuclear[0]); ++i ) {
26 number[ nuclear[ i ] ] = i + 1;
27 }
28
29 cin >> n >> k;
30 n2 = ( 1 << n );
31 n21 = n2 - 1;
32 for ( i = 0; i < n; ++i ) {
33 cin >> nuclearA[ i ];
34 numberA[ i ] = number[ nuclearA[ i ] ];
35 }
36 for ( i = 0; i < k; ++i ) {
37 cin >> nuclearB[ i ];
38 numberB[ i ] = number[ nuclearB[ i ] ];
39 }
40
41 memset( sum, 0, sizeof(sum) );
42 for ( s = 0; s < n2; ++s ) {
43 for ( j = 0; j < n; ++j ) {
44 if ( s & (1<<j) ) {
45 sum[ s ] += numberA[ j ];
46 }
47 }
48 }
49
50 memset( f, -1, sizeof(f) );
51 f[ 0 ] = 0;
52 for ( s = 0; s < n2; ++s ) {
53 i = f[ s ];
54 if ( (i==-1) || (i>=k) ) {
55 continue;
56 }
57 t = (s^n21);
58 for ( j = t; j >= 0; --j ) {
59 // nt = (j&t); // 超時
60 nt = j = (j&t);
61 if ( sum[ nt ] == numberB[ i ] ) {
62 f[ nt | s ] = i + 1;
63 p[ nt | s ] = s;
64 }
65 }
66 }
67
68 if ( f[ n21 ] < k ) {
69 cout << "NO" << endl;
70 }
71 else {
72 cout << "YES" << endl;
73 s = n21;
74 string str;
75 while ( s > 0 ) {
76 i = f[ s ] - 1;
77 t = p[ s ];
78 str = "";
79 for ( j = 0; j < n; ++j ) {
80 if ( ((s&(1<<j))!=0) && ((t&(1<<j))==0) ) {
81 str += nuclearA[ j ];
82 str += "+";
83 }
84 }
85 str.erase( str.length()-1 );
86 str += "->";
87 str += nuclearB[ i ];
88 cout << str << endl;
89 s = t;
90 }
91 }
92
93 return 0;
94 }
95