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請寫一個程式,把輸入的阿拉伯數字轉成羅馬數字輸出。

羅馬數字的規則請參考

https://zh.wikipedia.org/wiki/%E7%BD%97%E9%A9%AC%E6%95%B0%E5%AD%97

輸入說明:

會輸入一個不超過1000的正整數

輸出說明:

請將相對於輸入數字的羅馬數字輸出

輸入範例:

14

輸出範例:

XIV
[Exercise] Coding (C) - asked in 2017-1 程式設計(一)AD by (18k points)
ID: 34979 - Available when: Unlimited - Due to: Unlimited
| 1.1k views

6 Answers

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Hidden content!
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answered by (236 points)
0 0
prog.c: In function 'main':
prog.c:5:1: error: stray '\343' in program
 \xe3\x80\x80\xe3\x80\x80char a[4002],b[4002];
 ^
prog.c:5:2: error: stray '\200' in program
 \xe3\x80\x80\xe3\x80\x80char a[4002],b[4002];
  ^
prog.c:5:3: error: stray '\200' in program
 \xe3\x80\x80\xe3\x80\x80char a[4002],b[4002];
   ^
prog.c:5:4: error: stray '\343' in program
 \xe3\x80\x80\xe3\x80\x80char a[4002],b[4002];
    ^
prog.c:5:5: error: stray '\200' in program
 \xe3\x80\x80\xe3\x80\x80char a[4002],b[4002];
     ^
prog.c:5:6: error: stray '\200' in program
 \xe3\x80\x80\xe3\x80\x80char a[4002],b[4002];
      ^
prog.c:6:15: warning: format '%s' expects argument of type 'char *', but argument 2 has type 'char (*)[4002]' [-Wformat=]
 while(scanf("%s",&a)!=EOF)
               ^
prog.c:10:9: warning: format '%s' expects argument of type 'char *', but argument 2 has type 'char (*)[4002]' [-Wformat=]
 scanf("%s",&b);
         ^
prog.c:136:9: warning: implicit declaration of function 'abs' [-Wimplicit-function-declaration]
 int sum=abs(suma-sumb);
         ^~~
0 0
Case 0: Wrong output
Case 1: Wrong output
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Hidden content!
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answered by (215 points)
edited by
0 0
prog.c:1:1: error: expected identifier or '(' before '.' token
 .
 ^
0 0
Case 0: Wrong output
Case 1: Wrong output
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Hidden content!
* *** ** * *
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{
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answered by (100 points)
edited by
0 0
Case 0: Wrong output
Case 1: Wrong output
0 0
Case 0: Wrong output
Case 1: Wrong output
0 0
Case 0: Correct output
Case 1: Correct output
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Hidden content!
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answered by (174 points)
0 0
Case 0: Wrong output
Case 1: Wrong output
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Hidden content!
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return 0;
}
answered by (160 points)
0 0
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Case 1: Wrong output
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Hidden content!
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}
answered
0 0
Case 0: Correct output
Case 1: Correct output
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