uva 417 - Word Index(数位dp)
2014-05-22 21:34
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题目连接:uva
417 - Word Index
题目大意:按照题目中的要求,为字符串编号,现在给出字符串,问说编号为多少,注意字符串必须为递增的,否则编号为0。解题思路:其实就是算说比给定字符串小并且满足递增的串由多少个。dp[i][j]表示第i个位为j满足比给定字符串小并且满足递增的串。
dp[i][j]=∑k=0j−1dp[i−1][k].
注意每次要处理边界的情况,并且最后要加上自身串。并且在处理边界的时候dp[i][0]要被赋值为1,代表前i个为空的情况。
#include <cstdio> #include <cstring> #include <algorithm> using namespace std; const int N = 10; char str ; int dp [3*N]; int solve () { int len = strlen(str); if (len == 1) return str[0] - 'a' + 1;; for (int i = 1; i < len; i++) if (str[i] <= str[i-1]) return 0; int pre = str[0] - 'a' + 2; memset(dp, 0, sizeof(dp)); for (int i = 0; i + 'a' <= str[0]; i++) dp[1][i] = 1; for (int i = 1; i < len; i++) { for (int j = 0; j <= 26; j++) { for (int k = j+1; k <= 26; k++) dp[i+1][k] += dp[i][j]; } for (int j = pre; j + 'a' <= str[i]; j++) dp[i+1][j]++; pre = str[i] - 'a' + 2; dp[i+1][0]++; } int ans = 1; for (int i = 1; i <= 26; i++) ans += dp[len][i]; return ans; } int main () { while (scanf("%s", str) == 1) { printf("%d\n", solve()); } return 0; }
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