c语言答案56zhang.docx

1、c语言答案56zhangP1115.4#includevoidmain()inta,b,c;printf(请输入三个整数:);scanf(%d,%d,%d,&a,&b,&c);if(ab)if(bc)printf(max=%dn,c);elseprintf(max=%dn,b);elseif(ac)printf(max=%dn,c);elseprintf(max=%dn,a);方法二：#includevoidmain()inta,b,c,temp,max;printf(请输入三个整数:);scanf(%d,%d,%d,&a,&b,&c);temp=(ab)?a:b;max=(tempc)?te

2、mp:c;printf(最大数是%dn,max);5.5#includevoid main () int x,y; scanf(%d,&x); if(x1) y=x; printf(x=%d,y=x=%dn); else if (x10) y=2*x-1; printf(x=%d,y=2*x-1=%dn,x,y); else y=3*x-11; printf(x=%d,y=3*x-11=%dn,x,y); 5.6#includevoid main() float score; char grade; printf(请输入学生成绩); scanf(%f,&score); while(score1

3、00|score0) printf(n输入有误,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 %.1f,相应等级 %cn,score,grade); 5.7#includevoidmain()longintnum;intindi

4、v,ten,hundred,thousand,ten_thousand,place;printf(请输入一个整数：);scanf(%ld,&num);if(num9999)place=5;elseif(num999)place=4;elseif(num99)place=3;elseif(num9)place=2;elseplace=1;printf(place=%dn,place);printf(每位数字为：);ten_thousand=num/10000;thousand=(int)(num-ten_thousand*10000)/1000;hundred=(int)(num-ten_tho

5、usand*10000-thousand*1000)/100;ten=(int)(num-ten_thousand*10000-thousand*1000-hundred*100)/10;indiv=(int)(num-ten_thousand*10000-thousand*1000-hundred*100-ten*10);switch(place)case5:printf(%d,%d,%d,%d,%d,ten_thousand,thousand,hundred,ten,indiv);printf(n反序数字为：);printf(%d,%d,%d,%d,%d,indiv,ten,hundred

6、,thousand,ten_thousand);break;case4:printf(%d,%d,%d,%d,thousand,hundred,ten,indiv);printf(n反序数字为：);printf(%d,%d,%d,%d,indiv,ten,hundred,thousand);break;case3:printf(%d,%d,%d,hundred,ten,indiv);printf(n反序数字为：);printf(%d,%d,%d,indiv,ten,hundred);break;case2:printf(%d,%d,ten,indiv);printf(n反序数字为：);prin

7、tf(%d,%d,indiv,ten);break;case1:printf(%d,indiv);printf(n反序数字为：);printf(%d,indiv);break;5.8 #includevoidmain()longi;floatbonus,bon1,bon2,bon4,bon6,bon10;bon1=100000*0.1;bon2=bon1+100000*0.075;bon4=bon2+200000*0.05;bon6=bon4+200000*0.03;bon10=bon6+400000*0.015;scanf(%ld,&i);if(i=100000)bonus=i*0.1;el

8、seif(i=200000)bonus=bon1+(i-100000)*0.075;elseif(i=400000)bonus=bon2+(i-200000)*0.05;elseif(i=600000)bonus=bon6+(i-600000)*0.015;elsebonus=bon10+(i-1000000)*0.01;printf(奖金是%10.2f,bonus);法二：5.8#includevoidmain()longi;floatbonus,bon1,bon2,bon4,bon6,bon10;intc;bon1=bon2=bon4=bon6=bon10=printf(请输入利润i:);

9、scanf(%ld,&i);c=i/100000;if(c10)c=10;switch(c)case0:bonus=100000*0.1;break;case1:bonus=bon1+(i-100000)*0.075;break;case2:case3:bonus=bon2+(i-200000)*0.05;break;case4:case5:bonus=bon4+(i-400000)*0.03;break;case6:case7:case8:case9:bonus=bon6+(i-600000)*0.015;break;case10:bonus=bon10+(i-1000000)*0.01;p

10、rintf(奖金是%.2f,bonus);6.1#include void main()#includevoid main()int p,r,n,m,t;scanf(%d,%d,&m,&n);if(mn) t=m;m=n;n=t;p=m*n;while(n!=0) r=m%n; m=n; n=r;printf(最大公约数为：%dn,m);printf(最小公倍数为:%dn,p/m);6.2 #include void main() int a,n,i; long int sum=0,t; scanf(%d%d,&a,&n); i=1; while(i=n) t=a*(long)(pow(10,

11、i)-1)/9; sum+=t; i+; printf(sum=%ldn,sum); 6.3#includevoid main() int a,n,i=1,sn=0,t; printf(a,n=:); scanf(%d,%d,&a,&n);t=a; while(i=n) sn=sn+t; t=t*10+a; i+; printf(a+aa+aaa+.=%dn,sn);6.4 #includevoidmain()floats=0,t=1;intn;for(n=1;n=20;n+)t=t*n;s=s+t;printf(1!+2!+.+20!=%fn,s);6.5#includevoid main(

12、) int n1=100,n2=50,n3=10; float k; float s1=0,s2=0,s3=0; for(k=1;k=n1;k+) s1=s1+k; for(k=1;k=n2;k+) s2=s2+k*k; for(k=1;k=n3;k+) s3=s3+1/k; printf(总和=%8.2fn,s1+s2+s3);6.6# includevoid main() int i,j,k,n; printf(水仙花数是:); for(n=100;n1000;n+) i=n/100; j=n/10-i*10; k=n%10; if(n=i*i*i+j*j*j+k*k*k) printf(

13、%4d,n); 6.9#includevoid main( )float sn=100,hn=sn/2;int n;for(n=2;n=10;n+) sn=sn+2*hn; hn=hn/2;printf(第10次落地时共经过%f m.n,sn);printf(第10次反弹%f m.n,hn);6.10#includevoid main( )int day,x1,x2;day=9;x2=1;while(day0) x1=(x2+1)*2; x2=x1; day-;printf(total=%dn,x1);6.14#includevoid main( )int i,j,k;for(i=0;i=3;i+) for(j=0;j=2-i;j+) printf( ); for(k=0;k=2*i;k+) printf(*); printf(n);for(i=0;i=2;i+) for(j=0;j=i;j+) printf( ); for(k=0;k=4-2*i;k+) printf(*); printf(n);