結果

問題 No.463 魔法使いのすごろく🎲
ユーザー tails
提出日時 2016-12-14 01:05:28
言語 C90
(gcc 12.3.0)
結果
TLE  
実行時間 -
コード長 482 bytes
コンパイル時間 122 ms
コンパイル使用メモリ 21,248 KB
実行使用メモリ 13,760 KB
最終ジャッジ日時 2024-11-30 06:16:19
合計ジャッジ時間 20,794 ms
ジャッジサーバーID
(参考情報)
judge3 / judge4
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 34 TLE * 2
権限があれば一括ダウンロードができます
コンパイルメッセージ
main.c:3:1: warning: return type defaults to ‘int’ [-Wimplicit-int]
    3 | main(n,m,h,i,j,k){
      | ^~~~
main.c: In function ‘main’:
main.c:3:1: warning: type of ‘n’ defaults to ‘int’ [-Wimplicit-int]
main.c:3:1: warning: type of ‘m’ defaults to ‘int’ [-Wimplicit-int]
main.c:3:1: warning: type of ‘h’ defaults to ‘int’ [-Wimplicit-int]
main.c:3:1: warning: type of ‘i’ defaults to ‘int’ [-Wimplicit-int]
main.c:3:1: warning: type of ‘j’ defaults to ‘int’ [-Wimplicit-int]
main.c:3:1: warning: type of ‘k’ defaults to ‘int’ [-Wimplicit-int]
main.c:5:9: warning: implicit declaration of function ‘scanf’ [-Wimplicit-function-declaration]
    5 |         scanf("%d%d",&n,&m);
      |         ^~~~~
main.c:1:1: note: include ‘<stdio.h>’ or provide a declaration of ‘scanf’
  +++ |+#include <stdio.h>
    1 | double c[110],e0[110],e1[110];
main.c:5:9: warning: incompatible implicit declaration of built-in function ‘scanf’ [-Wbuiltin-declaration-mismatch]
    5 |         scanf("%d%d",&n,&m);
      |         ^~~~~
main.c:5:9: note: include ‘<stdio.h>’ or provide a declaration of ‘scanf’
main.c:31:9: warning: implicit declaration of function ‘printf’ [-Wimplicit-function-declaration]
   31 |         printf("%.10f\n",e0[1]);
      |         ^~~~~~
main.c:31:9: note: include ‘<stdio.h>’ or provide a declaration of ‘printf’
main.c:31:9: warning: incompatible implicit declaration of built-in function ‘printf’ [-Wbuiltin-declaration-mismatch]
main.c:31:9: note: include ‘<stdio.h>’ or provide a declaration of ‘printf’

ソースコード

diff #
プレゼンテーションモードにする

double c[110],e0[110],e1[110];
main(n,m,h,i,j,k){
scanf("%d%d",&n,&m);
for(i=2;i<n;++i){
scanf("%lf",&c[i]);
}
for(h=0;h<100000;++h){
for(i=n;--i;){
double e0i=0;
double e1i=0;
for(j=i+1;j<=i+m;++j){
k=j<n?j:n*2-j;
e0i+=e0[k]+c[k];
e1i+=e1[k]+c[k];
}
e0i/=m;
e1i/=m;
for(j=i+1;j<=i+m;++j){
k=j<n?j:n*2-j;
if(e0i>e1[k]+c[k]){
e0i=e1[k]+c[k];
}
}
e0[i]=e0i;
e1[i]=e1i;
}
}
printf("%.10f\n",e0[1]);
return 0;
}
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