UV 417-Word Index (Digital dp)

Question connection: Ultraviolet A 417-Word Index

According to the requirements in the question, it is a string number. Now we will give a string and ask how many strings are numbered. Note that the string must be incrementing; otherwise, the number is 0.

Solution: calculate the number of strings that are smaller than the given string and meet the increasing requirement. Dp [I] [j] indicates a string whose I-th position is j that is smaller than the given string and must be incremented.
Dp [I] [j] = Σ k = 0j? 1dp [I? 1] [k].
Pay attention to the situation of processing the boundary each time, and add its own strings at last. When processing the boundary, dp [I] [0] must be assigned a value of 1, which indicates that the first I is null.

#include 
  
   #include 
   
    #include using namespace std;const int N = 10;char str[N];int dp[N][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;}