The following program is my ACcepted code for UVA-11984.
It's a for everybody to learn and discuss.
If there is any mistake or comment, please let me know. :D
此乃UVA 11984 的AC code!
//題目來源 Problem link: http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=229&page=show_problem&problem=3135
#include<stdio.h>
int main()
{
int n;
double C, d;
scanf("%d", &n);
for(int count = 1; count <= n; count++)
{
scanf("%lf %lf", &C, &d);
C = C + (d * 5.0 / 9.0);
printf("Case %d: %.2lf\n", count, C);
}
return 0;
}
Please feel free to use it after adding this blog as an reference. (http://autekroy.blogspot.tw) If there is any mistake or comment, please let me know. :D
歡迎使用與分享任何內容,但請記得標示此部落格為出處。(http://autekroy.blogspot.tw/) 如果有發現任何的錯誤與建議請留言或跟我連絡。 : )
It's a for everybody to learn and discuss.
If there is any mistake or comment, please let me know. :D
此乃UVA 11984 的AC code!
歡迎一同討論學習,如有錯誤與任何建議請留言 : )
點這裡看題目 Click here to see this Problem!
d = F1 - F2
-> 因為d是差距,所以在兩個F相減下會把32抵銷
-> Because d is the difference, the 2 F will balance the 32
-> C = d * 5 / 9
//This program is for UVA 11984 A Change in Thermal Unitd = F1 - F2
-> 因為d是差距,所以在兩個F相減下會把32抵銷
-> Because d is the difference, the 2 F will balance the 32
-> C = d * 5 / 9
//題目來源 Problem link: http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=229&page=show_problem&problem=3135
#include<stdio.h>
int main()
{
int n;
double C, d;
scanf("%d", &n);
for(int count = 1; count <= n; count++)
{
scanf("%lf %lf", &C, &d);
C = C + (d * 5.0 / 9.0);
printf("Case %d: %.2lf\n", count, C);
}
return 0;
}
Please feel free to use it after adding this blog as an reference. (http://autekroy.blogspot.tw) If there is any mistake or comment, please let me know. :D
歡迎使用與分享任何內容,但請記得標示此部落格為出處。(http://autekroy.blogspot.tw/) 如果有發現任何的錯誤與建議請留言或跟我連絡。 : )
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