結果
問題 |
No.69 文字を自由に並び替え
|
ユーザー |
![]() |
提出日時 | 2025-03-04 09:38:37 |
言語 | C (gcc 13.3.0) |
結果 |
AC
|
実行時間 | 1 ms / 5,000 ms |
コード長 | 635 bytes |
コンパイル時間 | 518 ms |
コンパイル使用メモリ | 23,808 KB |
実行使用メモリ | 8,604 KB |
最終ジャッジ日時 | 2025-03-04 09:38:39 |
合計ジャッジ時間 | 1,650 ms |
ジャッジサーバーID (参考情報) |
judge5 / judge2 |
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ファイルパターン | 結果 |
---|---|
other | AC * 15 |
コンパイルメッセージ
main.c: In function ‘main’: main.c:5:22: warning: implicit declaration of function ‘malloc’ [-Wimplicit-function-declaration] 5 | char *strA = malloc(11); | ^~~~~~ main.c:2:1: note: include ‘<stdlib.h>’ or provide a declaration of ‘malloc’ 1 | #include <stdio.h> +++ |+#include <stdlib.h> 2 | main.c:5:22: warning: incompatible implicit declaration of built-in function ‘malloc’ [-Wbuiltin-declaration-mismatch] 5 | char *strA = malloc(11); | ^~~~~~ main.c:5:22: note: include ‘<stdlib.h>’ or provide a declaration of ‘malloc’ main.c:7:19: warning: implicit declaration of function ‘strlen’ [-Wimplicit-function-declaration] 7 | int lena =strlen(strA); | ^~~~~~ main.c:2:1: note: include ‘<string.h>’ or provide a declaration of ‘strlen’ 1 | #include <stdio.h> +++ |+#include <string.h> 2 | main.c:7:19: warning: incompatible implicit declaration of built-in function ‘strlen’ [-Wbuiltin-declaration-mismatch] 7 | int lena =strlen(strA); | ^~~~~~ main.c:7:19: note: include ‘<string.h>’ or provide a declaration of ‘strlen’ main.c:6:9: warning: ignoring return value of ‘scanf’ declared with attribute ‘warn_unused_result’ [-Wunused-result] 6 | scanf("%s",strA); | ^~~~~~~~~~~~~~~~ main.c:11:9: warning: ignoring return value of ‘scanf’ declared with attribute ‘warn_unused_result’ [-Wunused-result] 11 | scanf("%s",strB); | ^~~~~~~~~~~~~~~~
ソースコード
#include <stdio.h> int main(void){ // 文字列Aをスキャン・ポインタ char *strA = malloc(11); scanf("%s",strA); int lena =strlen(strA); char *a = strA; // 文字列Bをスキャン・ポインタ char *strB = malloc(11); scanf("%s",strB); int lenb = strlen(strB); char *b = strB; int count = 0; int ans = 0; for(int i = 0;i < lena;i++){ count = 0; for(int j = 0;j < lenb;j++){ //printf("A:%c B:%c\n",a[i],b[j]); if(a[i] == b[j]){ b[j] = '0'; count++; break; } } if(count == 0){ printf("NO"); break; } else{ ans++; } } if(ans == lenb){ printf("YES"); } }