C语言程序设计习题相关解答

谭浩强 C语言程序设计习题答案C语言程序设计习题参考答案 第一章 1.6 main() int a,b,c,max; printf(“input three numbers:n“); scanf(“%d,%d,%d“, max=a; if(maxb)?a:b; max=(ctemp)?c:temp; printf(“max=%d“,max); 4.5 main() int x,y; scanf(“%d“, if(x=0 else if(num999) place=4; else if(num99) place=3; else if(num9) place=2; else place=1; printf(“place=%dn“,place); ten_thousand=num/10000; thousand=(num-ten_thousand*10000)/1000; hundred=(num-ten_thousand*10000-thousand*1000)/100; ten=(num-ten_thousand*10000-thousand*1000-hundred*100)/10; indiv=num-ten_thousand*10000-thousand*1000-hundred*100-ten*10; switch(place) case 5:printf(“%d,%d,%d,%d,%dn“,ten_thousand,thousand,hundred,ten,indiv); printf(“%d,%d,%d,%d,%dn“,indiv,ten,hundred,thousand,ten_thousand); break; case 4:printf(“%d,%d,%d,%dn“,thousand,hundred,ten,indiv); printf(“%d,%d,%d,%dn“,indiv,ten,hundred,thousand); break; case 3:printf(“%d,%d,%dn“,hundred,ten,indiv); printf(“%d,%d,%dn“,indiv,ten,hundred); break; case 2:printf(“%d,%dn“,ten,indiv); printf(“%d,%dn“,indiv,ten); break; case 1:printf(“%dn“,indiv); printf(“%dn“,indiv); 4.8 main() long i; float bonus,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; 5 scanf(“%ld“, if(i10)branch=10; switch(branch) case 0:bonus=i*0.1;break; case 1:bonus=bon1+(i-100000)*