"C语言程序设计课后习题答案完整版及内控体系持续优化"

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(完整版)C语言程序设计课后习题答案.doc；(完整版)C语言程序设计课后习题答案.doc; In order to enhance the overall performance and effectiveness of our company, we are dedicated to continuously improve our internal control system. By doing so, we strive to achieve a more robust ability to manage and control our operations, as well as to optimize our business processes. The ultimate goal is to ensure that our processes run smoothly and that all responsibilities are properly allocated and executed. As part of our ongoing efforts, we are committed to further strengthening our internal controls. This includes playing the critical role of independent oversight in evaluating and ensuring compliance with third-party responsibilities. We recognize the importance of actively utilizing internal audit tools to detect potential management issues, streamline processes, and standardize related transactions. By doing so, we can effectively strengthen our operational capabilities and drive overall efficiency within the organization.

We will continue to improve the company's internal control system, and steady improvement in ability to manage and control, optimize business processes, to ensure smooth processes, responsibilities in place; to further strengthen internal controls, play a control post independent oversight role of evaluation complying with third-party responsibility; to actively make use of internal audit tools detect potential management, streamline, standardize related transactions, strengthening operations in accordance with law. Deepening the information management to ensure full communication "zero resistance". 　　To constantly perfect ERP, and BFS++, and PI, and MIS, and SCM, information system based construction, full integration information system, achieved information resources shared; to expand Portal system application of breadth and depth, play information system on enterprise of Assistant role; to perfect daily run maintenance operation of records, promote problem reasons

analysis and system handover; to strengthening BFS++, and ERP, and SCM, technology application of training, improve employees application information system of capacity and level. Humanistic care to ensure "zero." 　　To strengthening Humanities care, continues to foster company wind clear, and gas are, and heart Shun of culture atmosphere; strengthening love helped trapped, care difficult employees; carried out style activities, rich employees life; strengthening health and labour protection, organization career health medical, control career against; continues to implementation psychological warning prevention system, training employees health of character, and stable of mood and enterprising of attitude, created friendly fraternity of Humanities environment. To strengthen risk management, ensure that the business of "zero risk". To strengthened business plans management, will business business plans cover to all level, ensure the business can control in control; to close

concern financial, and coal electric linkage, and energy-saving scheduling, national policy trends, strengthening track, active should; to implementation State-owned assets method, further specification business financial management; to perfect risk tube control system, achieved risk recognition, and measure, and assessment, and report, and control feedback of closed ring management, improve risk prevention capacity. 　　To further standardize trading, and strive to achieve "according to law, standardize and fair." Innovation of performance management, to ensure that potential employees "zero fly". To strengthen performance management, process control, enhance employee evaluation and levels of effective communication to improve performance management. To further quantify and refine employee standards ... Work, full play party, and branch, and members in "five type Enterprise" construction in the of core role, and fighting fortress role and pioneer model role; to

continues to strengthening "four good" leadership construction, full play levels cadres in enterprise development in the

门急诊医技楼外脚手架采用悬挑式单排脚手架，基础回填后改为落地式双排脚手架，脚手架搭设高度为23m。病房医技楼外脚手架采用悬挑双排脚手架，分四次悬挑

of backbone backbone role; to full strengthening members youth work, full play youth employees in company development in the of force role; to improve independent Commission against corruption work level, strengthening on enterprise business key link of effectiveness monitored. , And maintain stability. To further strengthen publicity and education, improve the overall legal system. We must strengthen safety management, establish and improve the education, supervision, and evaluation as one of the traffic safety management mechanism. 　　To conscientiously sum up the Olympic security controls, promoting integrated management to a higher level, higher standards, a higher level of development. Employees, today is lunar calendar on December 24, the ox Bell is about to ring, at this time of year, we clearly feel the pulse of the XX power generation company to flourish, to more clearly hear XX power generation companies mature and symmetry breathing. Recalling past one another

across a railing, we are enthusiastic and full of confidence. Future development opportunities, we more exciting fight more spirited. 　　Employees, let us together across 2013 full of challenges and opportunities, to create a green, low-cost operation, full of humane care of a world-class power generation company and work hard! The occasion of the Spring Festival, my sincere wish that you and the families of the staff in the new year, good health, happy, happy

disc=b*b-4*a*c;

if(fabs(disc)<=1e-6) printf(“has tow eual roots:%8.4f\n”,-b/(2*a));

else if(fabs(disc)>1e-6)

{ x1=(-b+sqrt(disc))/(2*a);

x2=(-b-sqrt(disc))/(2*a);

printf(“has distinct real roots:%8.4f and %8.4f\n”,x1,x2);

}

else

{

realpart=-b/(2*a);

imagpart=sqrt(-disc)/(2*a);

printf(“has complex roots:\n”);

printf(“%8.4f+%8.4fi\n”,realpart,imagpart);

printf(“%8.4f-%8.4fi\n”,realpart,imagpart);

}

第五章

1. 求 100~200 间的全部素数。

解：

#include<stdio.h>

#include<math.h>

void main()

{

int m,i,n=0;

double k;

for(m=101;m<=200;m=m+2)

{

k=sqrt(m);

for(i=2;i<=k;i++)

if(m%i==0) break;

if(i>=k+1)

{

printf(“%d”,m); n=n+1;

}

if(n%10==0) printf(“\n”);

}

printf(“\n”);

}

2. 输入一行字符，分别统计出其中英文字母、空格、数字和其他字符的个数。

解：

#include<stdio.h>

void main()

{

continues to strengthening "four good" leadership construction, full play levels cadres in enterprise development in the

char c;

int i=0,j=0,k=0,l=0;

while((c=getchar())!=’\n’)

{

if(c>=’A’&&c<=’Z’||c>=’a’&&c<=’z’) i++;

else if(c>=’0’&&c<=’9’) j++;

else if(c==’ ’) k++;

else l++;

}

printf(“i=%d,j=%d,k=%d,l=%d\n”,i,j,k,l);

}

3. 输出所有“水仙花数”，所谓“水仙花数”是指一个三位数，其各位数字立方和等于该本

身。例如：153 是一个水仙花数，因为 153=1^3+5^3+3^3。

解：

#include<stdio.h>

#include<math.h>

void main()

{

int i,j,k,n;

printf(““水仙花”数是:”);

for(n=100;n<1000;n++)

{

i=n/100;

j=n/10-i*10;

k=n%10;

if(n==i*i*i+j*j*j+k*k*k) printf(“%4d”,n);

}

printf(“\n”);

}

4. 猴子吃桃问题。猴子第一天摘下若干个桃子，当即吃了一半，还不过瘾，又多吃了一个。

第二天早上又将剩下的桃子吃掉一半，又多吃一个。以后每天早上都吃了前一天剩下的一半

零一个。到第 10 天早上想再吃时，见只剩下一个桃子了。求第一天共摘多少桃子。

解：

#include<stdio.h>

void main()

{

int day,x1,x2;

day=9;

x2=1;

while(day>0)

{

x1=(x2+1)*2;

x2=x1;

day--;

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