• A. Spell Check


    time limit per test

    1 second

    memory limit per test

    256 megabytes

    input

    standard input

    output

    standard output

    Timur likes his name. As a spelling of his name, he allows any permutation of the letters of the name. For example, the following strings are valid spellings of his name: Timur, miurT, Trumi, mriTu. Note that the correct spelling must have uppercased T and lowercased other letters.

    Today he wrote string ss of length nn consisting only of uppercase or lowercase Latin letters. He asks you to check if ss is the correct spelling of his name.

    Input

    The first line of the input contains an integer tt (1≤t≤1031≤t≤103) — the number of test cases.

    The first line of each test case contains an integer nn (1≤n≤10)(1≤n≤10) — the length of string ss.

    The second line of each test case contains a string ss consisting of only uppercase or lowercase Latin characters.

    Output

    For each test case, output "YES" (without quotes) if ss satisfies the condition, and "NO" (without quotes) otherwise.

    You can output the answer in any case (for example, the strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive answer).

    Example

    input

    Copy

     
    

    10

    5

    Timur

    5

    miurT

    5

    Trumi

    5

    mriTu

    5

    timur

    4

    Timr

    6

    Timuur

    10

    codeforces

    10

    TimurTimur

    5

    TIMUR

    output

    Copy

    YES
    YES
    YES
    YES
    NO
    NO
    NO
    NO
    NO
    NO
    

    解题说明:此题是一道字符串题,首先判断长度对不对,然后再判断需要的字母是否都出现了即可。可以对字符串进行排序直接判断。

    1. #include
    2. #include
    3. int main()
    4. {
    5. int t;
    6. scanf("%d", &t);
    7. while (t--)
    8. {
    9. int n;
    10. scanf("%d", &n);
    11. char a[15], tem;
    12. scanf("%s", a);
    13. if (n == 5)
    14. {
    15. for (int i = 0; i<5; i++)
    16. {
    17. for (int j = i + 1; j<5; j++)
    18. {
    19. if (a[i]>a[j])
    20. {
    21. tem = a[i];
    22. a[i] = a[j];
    23. a[j] = tem;
    24. }
    25. }
    26. }
    27. if (a[0] == 'T'&&a[1] == 'i'&&a[2] == 'm'&&a[3] == 'r'&&a[4] == 'u')
    28. {
    29. printf("YES\n");
    30. }
    31. else
    32. {
    33. printf("NO\n");
    34. }
    35. }
    36. else
    37. {
    38. printf("NO\n");
    39. }
    40. }
    41. return 0;
    42. }

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  • 原文地址:https://blog.csdn.net/jj12345jj198999/article/details/127136380