首先,5不被视为列表的一部分,它是列表的计数.因此,它不应包括在计算中.
由于这是家庭作业,这里是伪代码.你的工作是首先理解伪代码(通过样本输入运行它)然后将其转换为C代码并尝试使其成功编译并运行(使用相同的样本输入).
我建议将样本输入"2 7 3"(两个项目,即7和3)作为一个良好的起点,因为它很小,总和将是10,最小3.
如果您已尝试超过一天,请将您的代码作为编辑发布到此问题中,我们会看到我们可以做些什么来帮助您.
get a number into quantity
set sum to zero
loop varying index from 1 to quantity
get a number into value
add value to sum
if index is 1
set smallest to value
else
if value is less than smallest
set smallest to value
endif
endif
endloop
output "The sum of the sequence of integers is: ", sum
output "The smallest of the integers entered is: ", smallest
Stack Overflow似乎分为三个阵营,那些只会给你代码的那些,那些会告诉你推卸并做自己的功课,以及那些像我一样宁愿看到你受过教育的人 - 当你点击时劳动力,我希望退休,所以你不会与我竞争:-).
在任何人在我的算法中选择漏洞之前,这是用于教育.我已经留下了至少一个问题,以帮助训练这个人 - 可能还有其他人,我会声称我故意将他们放在那里测试他:-).
更新:
罗伯特,在你已经评论过的(非常好的)尝试之后,这就是我修改你的代码来完成任务的方式(当然,不是我的手).您可以希望看到我的评论如何修改代码以达到此解决方案:
#includeint main (int argCount, char *argVal[]) { int i; // General purpose counter. int smallNum; // Holds the smallest number. int numSum; // Holds the sum of all numbers. int currentNum; // Holds the current number. int numCount; // Holds the count of numbers. // Get count of numbers and make sure it's in range 1 through 50. printf ("How many numbers will be entered (max 50)? "); scanf ("%d", &numCount); if ((numCount < 1) || (numCount > 50)) { printf ("Invalid count of %d.\n", numCount); return 1; } printf("\nEnter %d numbers then press enter after each entry:\n", numCount); // Set initial sum to zero, numbers will be added to this. numSum = 0; // Loop, getting and processing all numbers. for (i = 0; i < numCount; i++) { // Get the number. printf("%2d> ", i+1); scanf("%d", ¤tNum); // Add the number to sum. numSum += currentNum; // First number entered is always lowest. if (i == 0) { smallNum = currentNum; } else { // Replace if current is smaller. if (currentNum < smallNum) { smallNum = currentNum; } } } // Output results. printf ("The sum of the numbers is: %d\n", numSum); printf ("The smallest number is: %d\n", smallNum); return 0; }
以下是样本数据的输出:
pax> ./qq How many numbers will be entered (max 50)? 5 Enter 5 numbers then press enter after each entry: 1> 100 2> 350 3> 400 4> 550 5> 678 The sum of the numbers is: 2078 The smallest number is: 100 pax> ./qq How many numbers will be entered (max 50)? 5 Enter 5 numbers then press enter after each entry: 1> 40 2> 67 3> 9 4> 13 5> 98 The sum of the numbers is: 227 The smallest number is: 9 pax> ./qq How many numbers will be entered (max 50)? 0 Invalid count of 0. [fury]$ ./qq How many numbers will be entered (max 50)? 51 Invalid count of 51.
顺便说一下,请确保始终为代码添加注释.教育工作者喜欢那种东西.因此,开发人员必须在未来10年内尝试理解您的代码.
首先,5不被视为列表的一部分,它是列表的计数.因此,它不应包括在计算中.
由于这是家庭作业,这里是伪代码.你的工作是首先理解伪代码(通过样本输入运行它)然后将其转换为C代码并尝试使其成功编译并运行(使用相同的样本输入).
我建议将样本输入"2 7 3"(两个项目,即7和3)作为一个良好的起点,因为它很小,总和将是10,最小3.
如果您已尝试超过一天,请将您的代码作为编辑发布到此问题中,我们会看到我们可以做些什么来帮助您.
get a number into quantity
set sum to zero
loop varying index from 1 to quantity
get a number into value
add value to sum
if index is 1
set smallest to value
else
if value is less than smallest
set smallest to value
endif
endif
endloop
output "The sum of the sequence of integers is: ", sum
output "The smallest of the integers entered is: ", smallest
Stack Overflow似乎分为三个阵营,那些只会给你代码的那些,那些会告诉你推卸并做自己的功课,以及那些像我一样宁愿看到你受过教育的人 - 当你点击时劳动力,我希望退休,所以你不会与我竞争:-).
在任何人在我的算法中选择漏洞之前,这是用于教育.我已经留下了至少一个问题,以帮助训练这个人 - 可能还有其他人,我会声称我故意将他们放在那里测试他:-).
更新:
罗伯特,在你已经评论过的(非常好的)尝试之后,这就是我修改你的代码来完成任务的方式(当然,不是我的手).您可以希望看到我的评论如何修改代码以达到此解决方案:
#includeint main (int argCount, char *argVal[]) { int i; // General purpose counter. int smallNum; // Holds the smallest number. int numSum; // Holds the sum of all numbers. int currentNum; // Holds the current number. int numCount; // Holds the count of numbers. // Get count of numbers and make sure it's in range 1 through 50. printf ("How many numbers will be entered (max 50)? "); scanf ("%d", &numCount); if ((numCount < 1) || (numCount > 50)) { printf ("Invalid count of %d.\n", numCount); return 1; } printf("\nEnter %d numbers then press enter after each entry:\n", numCount); // Set initial sum to zero, numbers will be added to this. numSum = 0; // Loop, getting and processing all numbers. for (i = 0; i < numCount; i++) { // Get the number. printf("%2d> ", i+1); scanf("%d", ¤tNum); // Add the number to sum. numSum += currentNum; // First number entered is always lowest. if (i == 0) { smallNum = currentNum; } else { // Replace if current is smaller. if (currentNum < smallNum) { smallNum = currentNum; } } } // Output results. printf ("The sum of the numbers is: %d\n", numSum); printf ("The smallest number is: %d\n", smallNum); return 0; }
以下是样本数据的输出:
pax> ./qq How many numbers will be entered (max 50)? 5 Enter 5 numbers then press enter after each entry: 1> 100 2> 350 3> 400 4> 550 5> 678 The sum of the numbers is: 2078 The smallest number is: 100 pax> ./qq How many numbers will be entered (max 50)? 5 Enter 5 numbers then press enter after each entry: 1> 40 2> 67 3> 9 4> 13 5> 98 The sum of the numbers is: 227 The smallest number is: 9 pax> ./qq How many numbers will be entered (max 50)? 0 Invalid count of 0. [fury]$ ./qq How many numbers will be entered (max 50)? 51 Invalid count of 51.
顺便说一下,请确保始终为代码添加注释.教育工作者喜欢那种东西.因此,开发人员必须在未来10年内尝试理解您的代码.
读:
假设使用scanf读取的第一个整数指定剩余要输入的值的数量
所以它不是序列的一部分......
其余的,这是你的作业(和C ...)
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