Generalized Roman Numerals
64-bit integer IO format: %I64d Java class name: (Any)
The ancient Romans developed a terrible numbering system in which I, V, X, L and C stood for 1, 5, 10, 50 and 100, respectively. So XXXVII represents 37 (10+10+10+5+1+1). They typically wrote the numerals in non-increasing order. However, when a single Roman numeral is written before one that is larger, we subtract the smaller from the larger. So we can write IV and IX to represent 4 and 9 (subtracting 1), or XL and XC to represent 40 and 90 (subtracting 10). To represent 94, we would write XCIV. VIC is generally not considered a traditional Roman numeral, but we can interpret this as another representation of 94: VI is 6, so VIC is 100-6. In general, if we have two expressions a and b representing values v(a) and v(b), then we say that v(ab) is v(a) + v(b) if v(a) v(b), and v(b) ???? v(a) otherwise. Unfortunately, this generalization introduces some ambiguity, since dierent orders of evaluation may result in dierent values. For example, consider IVX: IV is 4 and X is 10, so by that reasoning IVX is 6. However, I is 1 and VX is 5, so this suggests that IVX is actually 4. To remedy this ambiguity, we allow the addition of parentheses. The question arises: for a given string of Roman numeral characters, how many dierent values can be obtained using dierent placements of parentheses?
Input
Each test case consists of a single string containing only the characters I, V, X, L and C. The length of
this string will be 50. A line containing of a single 0 will terminate input.
Output
For each test case, output all possible distinct values that can be represented by the string via the
addition of parentheses. Display these values in increasing order.
Sample Input
IVX
XIXIX
0
Sample Output
Case 1: 4 6
Case 2: 8 10 28 30 32
解题：区间dp...吓呆本宝宝 了5次幂的时间复杂度
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int maxn = 55; 4 unordered_set<int>dp[maxn][maxn]; 5 unordered_map<char,int>ump; 6 char str[maxn]; 7 void solve(int a,int b,int c) { 8 for(auto &A:dp[a][c]) 9 for(auto &B:dp[c+1][b]) { 10 if(A >= B) dp[a][b].insert(A + B); 11 else dp[a][b].insert(B - A); 12 } 13 } 14 int ret[10001000],tot; 15 int main() { 16 int cs = 1; 17 ump['I'] = 1; 18 ump['V'] = 5; 19 ump['X'] = 10; 20 ump['L'] = 50; 21 ump['C'] = 100; 22 while(~scanf("%s",str)) { 23 if(str[0] == '0') break; 24 for(int i = 0; i < maxn; ++i) 25 for(int j = 0; j < maxn; ++j) 26 dp[i][j].clear(); 27 int len = strlen(str); 28 for(int i = 0; i < len; ++i) 29 dp[i][i].insert(ump[str[i]]); 30 for(int i = 2; i <= len; ++i) 31 for(int j = 0; j + i <= len; ++j) 32 for(int k = j,t = j + i - 1; k < t; ++k) 33 solve(j,t,k); 34 tot = 0; 35 for(auto &it:dp[0][len-1]) ret[tot++] = it; 36 sort(ret,ret+tot); 37 printf("Case %d:",cs++); 38 for(int i = 0; i < tot; ++i) 39 printf(" %d",ret[i]); 40 putchar('\n'); 41 } 42 return 0; 43 }