本答案仅供参考（请不要随意更改）A类数学表达式（1—11）（N）1、#include<conio.h>#include<stdio.h>floatfun(intm,intn){doublet1=1,t2=1,t3=1,i;for(i=1;i<=m;i++)t1*=i;for(i=1;i<=n;i++)t2*=i;for(i=1;i<=m-n;i++)t3*=i;returnt1/(t2*t3);}main(){printf("P=%f\n",fun(12,8));}2#include<conio.h>#include<math.h>#include<stdio.h>floatfun(){doublex1=0.0,x0;x0=x1;x1=cos(x0);while(fabs(x0-x1)>=0.000001){x0=x1;x1=cos(x0);}returnx1;}main(){printf("Root=%f\n",fun());}3#include<conio.h>#include<stdio.h>doublefun(intn){doublef0=0,f1=1,fn;fn=f0+f1;while(fn<=n){f0=f1;f1=fn;fn=f0+f1;}returnfn;}main()/*主函数*/{printf("%f\n",fun(1000));}4#include<conio.h>#include<stdio.h>voidfun(float*sn,intn){floati,t=1;*sn=0;for(i=0;i<=n;i++){*sn+=t/(2*i+1);t=-t;}}main()/*主函数*/{intn=11;floats;fun(&s,n);printf("s=%fn=%d\n",s,n);}5#include<stdio.h>#include<math.h>doublefun(doublex[10]){doubles=0,sum=0;inti;for(i=0;i<10;i++)s=s+x[i];s=s/10;for(i=0;i<10;i++)sum+=(x[i]-s)*(x[i]-s);returnsqrt(sum/10);}main(){doubles,x[10]={95.0,89.0,76.0,65.0,88.0,72.0,85.0,81.0,90.0,56.0};inti;printf("\nTheoriginaldatais:\n");for(i=0;i<10;i++)printf("%6.1f",x[i]);printf("\n\n");s=fun(x);printf("s=%f\n\n",s);}6#include<conio.h>#include<math.h>#include<stdio.h>doublefun(intm){inti;doubles=0;for(i=1;i<=m;i++)s+=log(i);returnsqrt(s);}main(){printf("%f\n",fun(20));}7#include<conio.h>#include<stdio.h>#include<math.h>doublefun(doublex,intn){doubles=x+1,i,t=1;for(i=2;i<=n;i++){t*=i;s+=pow(x,i)/t;}returns;}main(){printf("%f\n",fun(0.3,10));}8#include<stdio.h>#include<math.h>doublefun(doubleeps){doubles=0,t=1,i,m=1;for(i=1;t/m>=eps;i++){s+=t/m;t=t*i;m=m*(2*i+1);}return2*s;}main(){doublex;printf("Inputeps:");scanf("%lf",&x);printf("\neps=%lf,PI=%lf\n",x,fun(x));}9#include<stdio.h>#include<math.h>doublefun(intn){doubles=0,i;for(i=1;i<=n;i++)s+=1.0/(i*(i+1));returns;}main(){printf("%f",