溫度華氏轉變攝氏
#include <stdio.h>
#include <stdlib.h>
enum x {A,B,C,D,E}
int main(void)
{
int a=73,b=85,c=66
{
if (a>=90)
printf("a=A等級!!\n")
else if (a>=80)
printf("73分=B等級!!\n")
else if (a>=70)
printf("73分=C等級!!\n")
else if (a>=60)
printf("73分=D等級!!\n")
else if (a<60)
printf("73分=E等級!!\n")
}
{
if (b>=90)
printf("b=A等級!!\n")
else if (b>=80)
printf("85分=B等級!!\n")
else if (b>=70)
printf("85分=C等級!!\n")
else if (b>=60)
printf("85分=D等級!!\n")
else if (b<60)
printf("85分=E等級!!\n")
}
{
if (c>=90)
printf("c=A等級!!\n")
else if (c>=80)
printf("66分=B等級!!\n")
else if (c>=70)
printf("66分=C等級!!\n")
else if (c>=60)
printf("66分=D等級!!\n")
else if (c<60)
printf("66分=E等級!!\n")
}
system("pause")
return 0
}
溫度華氏轉變攝氏
#include <stdio.h>
#include <stdlib.h>
enum x {A,B,C,D,E}
int main(void)
{
int a=73,b=85,c=66
{
if (a>=90)
printf("a=A等級!!\n")
else if (a>=80)
printf("73分=B等級!!\n")
else if (a>=70)
printf("73分=C等級!!\n")
else if (a>=60)
printf("73分=D等級!!\n")
else if (a<60)
printf("73分=E等級!!\n")
}
{
if (b>=90)
printf("b=A等級!!\n")
else if (b>=80)
printf("85分=B等級!!\n")
else if (b>=70)
printf("85分=C等級!!\n")
else if (b>=60)
printf("85分=D等級!!\n")
else if (b<60)
printf("85分=E等級!!\n")
}
{
if (c>=90)
printf("c=A等級!!\n")
else if (c>=80)
printf("66分=B等級!!\n")
else if (c>=70)
printf("66分=C等級!!\n")
else if (c>=60)
printf("66分=D等級!!\n")
else if (c<60)
printf("66分=E等級!!\n")
}
system("pause")
return 0
}
We have a group of N items (represented by integers from 1 to N), and we know that there is some total order defined for these items. You may assume that no two elements will be equal (for all a, b: a<b or b<a). However, it is expensive to compare two items. Your task is to make a number of comparisons, and then output the sorted order. The cost of determining if a < b is given by the bth integer of element a of costs (space delimited), which is the same as the ath integer of element b. Naturally, you will be judged on the total cost of the comparisons you make before outputting the sorted order. If your order is incorrect, you will receive a 0. Otherwise, your score will be opt/cost, where opt is the best cost anyone has achieved and cost is the total cost of the comparisons you make (so your score for a test case will be between 0 and 1). Your score for the problem will simply be the sum of your scores for the individual test cases.
1. 下列說法正確的是 （ ）
A. Java語言不區分大小寫
B. Java程序以類為基本單位
C. JVM為Java虛擬機JVM的英文縮寫
D. 運行Java程序需要先安裝JDK
2. 下列說法中錯誤的是 ( )
A. Java語言是編譯執行的
B. Java中使用了多進程技術
C. Java的單行注視以//開頭
D. Java語言具有很高的安全性
3. 下面不屬于Java語言特點的一項是( )
A. 安全性
B. 分布式
C. 移植性
D. 編譯執行
4. 下列語句中，正確的項是 ( )
A . int $e,a,b=10
B. char c,d=’a’
C. float e=0.0d
D. double c=0.0f
