1、天津理工大学c语言上机报告4实验四 (2009-9-24)一、 实验名称:程序控制结构。二、 实验目的:(1) 掌握C语言基本程序结构(顺序、分支和循环)及其应用; (2) 结合程序理解一些常用的算法(穷举、迭代和递推等);(3) 进一步熟悉Visual C+6.0开发环境下调试程序的基本步骤。三、 实验项目(1) 源程序:#include int main() int x; int y; /输入(Input) printf(Enter an integer: n); scanf(%d, &x); /处理(Process) if(x1) y=x; else if(x6) y=3*x-2; el
2、se y=4*x-8; /输出(Output) printf(y=%dn, y); return 0;(2) 采用if语句时,源程序为:#include int main() int score; /输入(Input) printf(Enter the score: n); scanf(%d, &score); /处理(Process)和输出(Output) if(score100|score=90) printf(Excellentn); else if(score=80) printf(Goodn); else if(score=70) printf(Middlen); else if(s
3、core=60) printf(Passedn); else printf(Failedn); return 0;采用swtich语句时,源程序为: #include int main() int score; /输入(Input) printf(Enter the score: n); scanf(%d, &score); /处理(Process)和输出(Output) switch(score/10) case 10: case 9: printf(Excellentn); break; case 8: printf(Goodn); break; case 7: printf(Middle
4、n); break; case 6: printf(Passedn); break; case 5: case 4: case 3: case 2: case 1: case 0: printf(Failedn); break; default: printf(Illegaln); break; return 0;(3) 源程序为:#include int main() int data1, data2; char op; double result; /输入(Input) printf(Enter the expression: data1 op data2n); scanf(%d%c%d,
5、 &data1, &op, &data2); /处理(Process) switch(op) case +: result = data1+data2; break; case -: result = data1-data2; break; case *: result = data1*data2; break; case /: if(data2!=0) result = data1/data2; else printf(Divided by zero!n); return 1; break; /输出(Output) printf(%d%c%d=%7.2fn, data1, op, data2
6、, result); return 0;(4) 源程序为:int main() int line; int loop; /处理(Process)和输出(Output) line=0; for(loop=0; loop=1000; loop=loop+1) if(loop%13=0 & loop%7=0) printf(%d , loop); +line; if(line%10=0) / 每行打印10个数 printf(n); printf(n); return 0;(5) 源程序为:#include int main() int loop; int num; int den; int tmp;
7、 double sum; /处理(Process) sum=0; num=2; den=1; for(loop=1; loop=20; loop=loop+1) sum += num*1.0/den; tmp = num; num = num+den; den = tmp; /输出(Output) printf(sum=%7.2fn, sum); return 0;(6) 源程序:#include int main() int loop; int num; int n_even; int sum; /输入和处理(Process) sum=0; n_even=0; for(loop=1; loo
8、p=10; loop=loop+1) scanf(%d, &num); if(num%2=0) sum += num; +n_even; /输出(Output) printf(偶数之和sum=%dn, sum); printf(偶数的平均数ave=%7.2fn, sum*10/n_even); return 0;(7) 源程序:#include int main() int loop; double fn; double fact; double sum; /处理(Process) sum=0; fn=1.0; loop=1; fact=1; while(fn=1e-6) sum += fn; fact *= loop; fn = 1/fact; loop = loop+1; /输出(Output) printf(sum=%7.5fn, sum); return 0;
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