_______________)!=eof&&i{if(c>=’0’&&c<=’9’||c>=’a’&&c<=’f’||c>=’A’&&c<=’F’){isdigit=yes;t[i++]=___________;}elseif(isdigit==yes){isdigit=no;t[i]=’\0’;printf(“%d”,______________);i=0;}}}inthtoi(char*s){inti,n;n=0;for(i=0;s[i]!=’\0’;i++){if(s[i]>=’0’&&s[i]<=’9’)n=n*16+s[i]-______________;if(s[i]>=’a’&&s[i]<=’f’)n=n*16+s[i]-‘a’+10;if(s[i]>=’A’&&s[i]<=’F’)n=n*16+s[i]-‘A’+10;}return___________;}试题二:以下程序的功能是应用近似公式计算e计算每项分母的值,共取nmax项之和作为e的近似值。近似公式:e=1+x+x/2!+x/3!+……+x/(nmax-1)!floatf2(intn){if(n==1)return1;elsereturn____________________;}floatf1(floatx,intn){inti;floatj=______________;for(i=1;___________;i++)j=j*x;returnj;}#definenmax20main(){floatx,exp=1.0;intn;printf(“inputxvalue:”);scanf(“%f”,&x);printf(“%f\n”,x);for(n=1;nexp=______________;printf(“x=%f,exp(x)=%f\n”,x,_______________);}答案:第一卷:
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