a=23 b=56 i =-23 if i > 0 thena=23b=56i =-23if i > 0 thenc = a + belsec = a – bend ifc的值是()A .79 B.-33 C.33 D.-79

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a=23 b=56 i =-23 if i > 0 thena=23b=56i =-23if i > 0 thenc = a + belsec = a – bend ifc的值是()A .79 B.-33 C.33 D.-79
a=23 b=56 i =-23 if i > 0 then
a=23
b=56
i =-23
if i > 0 then
c = a + b
else
c = a – b
end if
c的值是()
A .79 B.-33 C.33 D.-79

a=23 b=56 i =-23 if i > 0 thena=23b=56i =-23if i > 0 thenc = a + belsec = a – bend ifc的值是()A .79 B.-33 C.33 D.-79
答案是B.

if(a[i]>=b[i]&&a[i] a=23 b=56 i =-23 if i > 0 thena=23b=56i =-23if i > 0 thenc = a + belsec = a – bend ifc的值是()A .79 B.-33 C.33 D.-79 if(s[i]>=‘a’&&s[i] if(isP(a[i])) b[cnt++]=a[i];什么意思?我这些都不怎么懂. IF公式的运用=IF(F10=Vc,28,IF(F10=Vb,28,IF(F10=Va,28,IF(F10=Ⅳc,25,IF(F10=Ⅳb,25,IF(F10=Ⅳa,25,IF(F10=Ⅲb,23,IF(F10=Ⅲa,12,5)))))))) .F10是单元格 想再加IF(F10=Ⅱb,5,). main() { int a,b,i,min,n; scanf(%d%d,&a,&b); min=a; if(b/a!=n&&n>1) { for(i=min/2;i>0;i--) if(a%i请问怎么求这两个数的最大公约数 a1){for(i=min/2;i>0;i--)if(a%i==0&&b%i==0) break;}else if(b/a==n&&n>1)i=a;printf(%d ,i);}程序应该 c++,用户输入10个数,要求输出出第几个数是最大值,#includeusing namespace std;int main(){ int i,a[10],b,c,max,min; couta[0]; max=min=a[0]; for(i=1;i>a[i]; if(maxa[i])min=a[i]; } if(a[i]=max)i=b; if(a[i]=min)i=c; cout 以下是无限循环的语句是 并分析A for(i=1; ;){if(i++%2==0)continue;if(i++%3==0)break;B i=32767do{if(i int x=1,y=1,i=0;则下列选项错误的是()a if(xy)i--,b if(x||y)i++,c(x+y)i++,d if(x=y)i-- a=1 b:if a if(i!='x' =IF(A2>90,A ,IF(A2>80,B ,C )) “=IF(A2>90,A,IF(A2>80,B,C))”, [求助] 函数突破了7层,如何换种方法表示 ?=IF(X3>0,A,IF((W3+V3)>0,B,IF((T3+U3)>0,C,IF(S3>0,D,IF(R3>0,E,IF(Q3>0,F,IF(O3>0,G,IF(N3>0,H,IF(M3>0,I,IF(L3>0,J,IF(K3>0,K,IF(J3>0,L,IF(I3>0,M,)))))))))))))))))函数突 if ((a>=4'b0111) && (b if(a[i]==a.length-1)是什么意思 1.已知 int i,x,y;在下列选项中错误的是( ).A:if(x == y)i++; B:if(x = y)i--; C:if( xy =IF(A1=a,1,IF(A2=b,2,IF(A3=c,3,IF(A4=d,4,IF(A5=e,5,IF(A6=f,6,IF(A7=g,7,IF(A8=h,8,if(A9=i,9,if(a10=j,10,if(a11=k,11,if(a12=l,12,if(a13=m,13,if(a14=n,14,if(a15=o,15,if(a16=p,16,if(a17=q,17,if(a18=r,18,19)))))))))))))