1）计算1～100之间所有含8的数的和。#include<stdio.h>main(){inti=1,sum=0,a,b;for(i=1;i<=100;i++){if(i<10)if(i==8)sum=sum+i;elsecontinue;if(i>=10&&i<100){a=i/10;b=i%10;if(a==8||b==8)sum=sum+i;elsecontinue;}}printf("sum=%d",sum);}2）用下列近似公式计算e值，误差应小于10-5。…方法1#include<stdio.h>#include<math.h>main(){intfac(intn);doublee=1.0,t=1.0;intn=1;while(fabs(t)>=1e-5){t=1.0/fac(n);//printf("%ld\n",fac(n));e=e+t;n++;}printf("e=%f\n",e);}intfac(intn){inti,z=1;for(i=1;i<=n;i++)z=z*i;returnz;}方法2#include<stdio.h>#include<math.h>main(){intfac(intn);doublee=1.0,t=1.0;intn=1;while(fabs(t)>=1e-5){t=1.0/fac(n);e=e+t;n++;}printf("e=%f\n",e);}intfac(intn){longintz=1;if(n==1)return1;else{z=n*fac(n-1);returnz;}}编写程序验证下列结论：任何一个自然数n的立方都等于n个连续奇数之和。例如：13=1；23=3+5；33=7+9+11;43=13+15+17+19。要求：程序对每个输入的自然数计算并输出相应的连续奇数，直到输入的自然数为0时止。方法1#include<stdio.h>#include<math.h>main(){intn,i,b,c=0,j,k,l;printf("pleaseinput:");scanf("%d",&n);while(n!=0){b=pow(n,3);printf("cubeof%dis%d=",n,b);for(i=1;i<=b;i=i+2){c=0;l=i;for(j=1;j<=n;j++){c=c+l;l=l+2;}//c用来存储n个连续奇数的和//printf("\nc=%d,l=%d\n",c,l);if(b==c){l=l-2;for(k=1;k<=n;k++){if(k<n)printf("%d+",l);elseprintf("%d\n",l);l=l-2;}break;}}printf("\npleaseinput:");scanf("%d",&n);//printf("\na=%d\n",a);}方法2#include<stdio.h>#include<math.h>main(){inta,b,c=0,j,l;printf("pleaseinput:");scanf("%d",&a);while(a!=0){b=pow(a,3);printf("cubeof%dis%d=",a,b);{c=0;l=a*a-a+1;for(j=1;j<=a;j++){if(j==a)printf("%d\n",l);elseprintf("%d+",l);c=c+l;l=l+2;}//c用来存储三个连续奇数的和}printf("\npleaseinput:");scanf("%d",&a);}}