Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

Problem E: Lovely Hint

Jay and Kay decide to play a game. They write an English sentence on a piece of paper. The objective of the game is to pick out some of the written alphabets, rearrange them, and form the longest "lovely string".

The definitions of "lovely string" are simple:

1. It consists of upper case alphabets only, and it cannot be an empty string;
2. All alphabets are assigned a value. For example, A is assigned to be 1, B is assigned to be 2...etc;
3. For any substring of two alphabets, suppose the value of the first alphabet is V₁ and that of the second alphabet is V₂. Then the following inequality must hold: 5 × V₁ ≤ 4 × V₂.

Jay asks you to help him. However, as he does not want to lose the fun of the game, he asks you to give him a hint only. Write a program to give the hint.

Input

The input contains an integer N (1 ≤ N ≤ 30). Each of the next N lines consists of a string S. The string can be as long as 250 characters, and will not contain lower case alphabets.

Output

For each case, print the length of the longest lovely string that can be formed using the alphabets, followed by the number of ways to achieve this length.

Sample Input

2
HELLO.
I AM JAY.

Sample Output

4 1
4 2

Explanation

For the first case, we can get the lovely string EHLO.
For the second case, we can get AIMY and AJMY.

---

不用管重複的，只要選有出現過的大寫字母即可，
然後由小排到大，根據它給定的條件做 LIS 就好。

#include <stdio.h>
#include <algorithm>
using namespace std;
int main() {
    char cmd[300];
    int dp[300], way[300];
    gets(cmd);
    while(gets(cmd)) {
        int cnt[26] = {}, i, j;
        for(i = 0; cmd[i]; i++) {
            if(cmd[i] >= 'A' && cmd[i] <= 'Z')
                cnt[cmd[i]-'A']++;
        }
        for(i = 0, j = 0; i < 26; i++)
            if(cnt[i])
            cmd[j++] = i+1;
        int n = j;
        int mx = 0, mxway;
        for(i = 0; i < n; i++) {
            dp[i] = 1, way[i] = 0;
            for(j = 0; j < i; j++)
                if(cmd[j]*5 <= cmd[i]*4)
                    dp[i] = max(dp[i], dp[j]+1);
            for(j = 0; j < i; j++)
                if(cmd[j]*5 <= cmd[i]*4 && dp[i] == dp[j]+1)
                    way[i] += way[j];
            if(dp[i] == 1)  way[i] = 1;
            if(dp[i] > mx)  mx = dp[i], mxway = 0;
            if(dp[i] == mx) mxway += way[i];
        }
        printf("%d %d\n", mx, mxway);
    }
    return 0;
}