int x = 1;
int sum = 0;//和,初始化為0
while (x <= 30)//循環條件
{
sum += x;
x++;
}
Console.Write("1-30奇數的和:" + (sum - (5 * 3)) / 2);
int sum = 0;//和,初始化為0
while (x <= 30)//循環條件
{
sum += x;
x++;
}
Console.Write("1-30奇數的和:" + (sum - (5 * 3)) / 2);
2016-03-25
int x = 1;
int sum = 0;//和,初始化為0
while (x <= 30)//循環條件
{
if (x%2!=0){//篩選條件
sum += x;
}
x++;
}
Console.Write("1-30奇數的和:"+sum);
int sum = 0;//和,初始化為0
while (x <= 30)//循環條件
{
if (x%2!=0){//篩選條件
sum += x;
}
x++;
}
Console.Write("1-30奇數的和:"+sum);
2016-03-25
int x = 1;
int sum = 0;//和,初始化為0
while (x <= 30)//循環條件
{
if (x%2!=0)//篩選條件
sum += x;
x++;
}
Console.Write("1-30奇數的和:"+sum);
int sum = 0;//和,初始化為0
while (x <= 30)//循環條件
{
if (x%2!=0)//篩選條件
sum += x;
x++;
}
Console.Write("1-30奇數的和:"+sum);
2016-03-24
int[] score = { 90,79,67,88,60,99,100,65};
string[] name = {"Amy","Tom","Jack","Alice","Mark","Sun","King","Davie" };
int avg, sum = 0;
for (int i = 0; i < score.Length; i++)
{
sum += score[i];
}
string[] name = {"Amy","Tom","Jack","Alice","Mark","Sun","King","Davie" };
int avg, sum = 0;
for (int i = 0; i < score.Length; i++)
{
sum += score[i];
}
題目要求:輸出結果為4
即 x-y的值要求為4,程序已經給出了x的值為 x/=0.5,即x=4。那么y的值只能是0.
以上,可知 使y取余或者減2都可以滿足.
即 x-y的值要求為4,程序已經給出了x的值為 x/=0.5,即x=4。那么y的值只能是0.
以上,可知 使y取余或者減2都可以滿足.
2016-03-24