C程序设计(第四版)谭浩强第4章习题答案.doc

第四章4.4#includeint main()int a,b,c,t;scanf(%d,%d,%d,&a,&b,&c);if(ab) t=a; a=b; b=t;if(ac) t=a; a=c; c=t;if(bc) t=b; b=c; c=t;printf(%dn,c);return 0; 4.5#include#includeint main()int a,x1,x2;scanf(%d,&a);if(a=1000)printf(请重新输入一个小于1000的值a:);scanf(%d,&a);else x1=sqrt(a); x2=-sqrt(a);printf(x1=%d,x2=%dn,x1,x2);return 0;4.6#include#includeint main()float x,y;scanf(%f,&x);if(x1) y=x;else if(x10) y=2*x-1; else y=3*x-11;printf(y=%fn,y);return 0;4.7#includeint main()int x,y;printf(enter x:);scanf(%d,&x);if(x0) y=-1;else if(x=0) y=0; else y=1;printf(y=%dn,y);return 0;4.8#includeint main()float score; char grade; printf(请输入学生成绩:); scanf(%f,&score); while(score100|score0) printf(输入错误，请重输); scanf(%f,&score); switch(int)(score/10) case 10: case 9:grade=A;break; case 8:grade=B;break; case 7:grade=C;break; case 6:grade=D;break; case 5: case 4: case 3: case 2: case 1: case 0:grade=E;printf(成绩是%6.2f,相应的等级是%cn,score,grade);return 0;4.9#include int main()int num, g,s,b,q,w,place;printf(输入整数：);scanf(%d,&num);if(num9999) place=5;else if(num999) place=4;else if(num99) place=3;else if(num9) place=2;else place=1;printf(位数：%dn,place);printf(每一位数字：);w=num/10000;q=(int)(num-w*10000)/1000;b=(int)(num-w*10000-q*1000)/100;s=(int)(num-w*10000-q*1000-b*100)/10;g=num-w*10000-q*1000-b*100-s*10;switch(place)case 5:printf(%d,%d,%d,%d,%dn,w,q,b,s,g); printf(反序数字为：%d%d%d%d%dn,g,s,b,q,w);break; case 4:printf(%d,%d,%d,%dn,q,b,s,g); printf(反序数字为：%d%d%d%dn%,g,s,b,q);break; case 3:printf(%d,%d,%dn,b,s,g); printf(反序数字为：%d%d%dn%,g,s,b);break; case 2:printf(%d,%dn,s,g); printf(反序数字为：%d%dn,g,s);break; case 1:printf(%dn,g); printf(反序数字为：%dn,g);break;return 0;4.10#includeint main()double ti,i1,i2,i4,i6,i10;int i;printf(输入利润:);scanf(%d,&i);i1=100000*0.1;i2=i1+100000*0.075;i4=i2+200000*0.05;i6=i4+200000*0.03;i10=i6+400000*0.015;if(i=100000) ti=i*0.1;else if(i=200000) ti=i1+(i-100000)*0.075;else if(i=400000) ti=i2+(i-200000)*0.05;else if(i=600000) ti=i4+(i-400000)*0.03;else if(i=1000000) ti=i6+(i-600000)*0.015;else ti=i10+(i-1000000)*0.01;printf(应发奖金总数为%10.2fn,ti);return 0;4.10#includeint main()double ti,i1,i2,i4,i6,i10;int i,t;printf(输入利润:);scanf(%d,&i);i1=100000*0.1;i2=i1+100000*0.075;i4=i2+200000*0.05;i6=i4+200000*0.03;i10=i6+400000*0.015;t=i/100000;if(t10) t=10;switch(t) case 0:ti=i*0.1;break;case 1:ti=i1+(i-100000)*0.075;break;case 2:case 3:ti=i2+(i-200000)*0.05;break;case 4:case 5:ti=i4+(i-400000)*0.03;break;case 6:case 7:case 8:case 9:ti=i6+(i-600000)*0.015;break;case 10:ti=i10+(i-1000000)*0.01;printf(应发奖金总数为%10.2fn,ti);return 0;4.11#includeint main()int a,b,c,d,t;scanf(%d,%d,%d,%d,&a,&b,&c,&d);if(ab) t=a; a=b; b=t;if(ac) t=a; a=c; c=t;if(ad) t=a; a=d; d=t;if(bc) t=b; b=c; c=t;if(bd) t=b; b=d; d=t;if(cd) t=c; c=d; d=t; printf(%d,%d,%d,%dn,a,b,c,d);return 0;4.12#includeint main()float x,y,d1,d2,d3,d4,x1=2,y1=2,x2=-2,y2=2,x3=-2,y3=-2,x4=2,y4=-2;int h=10;printf(输入坐标(X,Y):);scanf(%f,%f,&x,&y);d1=(x-x1)*(x-x1)+(y-y1