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XZG_AverSumofEveryCourse流程图

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XZG_AverSumofEveryCourse流程图
xuzigui 2 years ago
parent 1bcd4f8f5d
commit ed2925711d
2 changed files with 145 additions and 1 deletions
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AverSumofEveryCourse.svg
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<!-- Created by Flowgorithm 3.4.2 (http://flowgorithm.org) -->
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<text x="496" y="142" fill="#000000" font-family="Arial" font-size="32pt" text-anchor="middle" direction="ltr">
<tspan x="496" dy="62" unicode-bidi="embed">AverSumofEveryCourse</tspan>
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<text x="511" y="390" fill="#000000" font-family="Arial" font-size="32pt" text-anchor="middle" direction="ltr">
<tspan x="511" dy="62" unicode-bidi="embed">Integer i</tspan>
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<tspan x="511" dy="62" unicode-bidi="embed">Integer j</tspan>
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<tspan x="511" dy="62" unicode-bidi="embed">Real sum[COURSE_NUM]</tspan>
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<tspan x="1178" dy="62" unicode-bidi="embed">sum[j] = 0</tspan>
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<tspan x="1736" dy="62" unicode-bidi="embed">sum[j] = sum[j]+stu[i].score</tspan>
<tspan x="1736" dy="62" unicode-bidi="embed">[j]</tspan>
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<tspan x="1178" dy="62" unicode-bidi="embed">i = 0 to n-1</tspan>
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<tspan x="1519" dy="62" unicode-bidi="embed">Next</tspan>
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<tspan x="1069" dy="62" unicode-bidi="embed">Done</tspan>
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<tspan x="1178" dy="62" unicode-bidi="embed">aver[j] = sum[i]/i</tspan>
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<tspan x="1178" dy="62" unicode-bidi="embed">Output &quot;course %d: sum =</tspan>
<tspan x="1178" dy="62" unicode-bidi="embed"> %.0f,aver = %.0f\n&quot;, j + 1,</tspan>
<tspan x="1178" dy="62" unicode-bidi="embed"> sum[j], aver[j]</tspan>
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<tspan x="496" dy="62" unicode-bidi="embed">j = 0 to m-1</tspan>
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<tspan x="837" dy="62" unicode-bidi="embed">Next</tspan>
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<tspan x="387" dy="62" unicode-bidi="embed">Done</tspan>
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README.md
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@ -423,12 +423,19 @@ Step 2：利用循环，输入学生的学号，成绩和各科成绩
Step 1:利用for循环给 sum 赋初值为 0
Step 2:利用第二重循环重新计算每个学生的总分
Step 3:计算每个学生的平均分
Step 4:将结果按照舒徐，总分和平均分的顺序进行输出
Step 4:将结果按照顺序，总分和平均分的顺序进行输出
![计算每个学生各门课程的总分和平均分](AverSumofEveryStudent.svg)
#### AverSumofEveryCourse
step 1:利用for循环给 sum 赋初值为 0
Step 2:利用第二重循环重新计算每门课程的总分
Step 3:计算每门课程的平均分
tep 4:将结果按照顺序，总分和平均分的顺序进行输出
![计算每门课程的总分和平均分](AverSumofEveryCourse.svg)
#### SortbyScore
step 1:利用循环将i赋值给k
step 2:用循环从i+1到n-1比较,将较小的赋给k

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