1.A D D C(A因为设置一个初始值为0)D(改变数的类型)D(分母的数值每次变大2)D(因为sum的类型是double)
2.D A B C(D因为设置a【i】为整形)A(数组的初始值为0)B可以是输入有10个值C(获取最大值)
3.
#include<stdio.h> #include<math.h> int main(void) { double x,y; printf("The Enter x:"); scanf("%lf",&x); if(x<-2){ y=x*x; } else if(-2<=x&&x<=2){ y=2+x; } else{ y=sqrt(x*x+x+1); } printf("y=%.2f",y); return 0; }
4.
#include<stdio.h> int main(void) { int count,i,n,number,good;/*count表示不及格的人数,number表示合格人数,good表示优秀人数*/ double grade; printf("The Enter n:"); scanf("%d",&n); count=0; number=0; good=0; for(i=1;i<=n;i++){ printf("The Enter grade:"); scanf("%lf",&grade); if(grade<60&&grade>0) count++; if(grade>60&&grade<84) number++; if(grade>84) good++; if(grade<0) printf("The enter is wrong "); } printf("The Number of failure=%d ",count); printf("The Number of pass=%d ",number); printf("The number of good=%d ",good); return 0; }