c程,编写一个函数将任意一个正整数N的立方分解成n个连续的奇数和

intfun(intn){inta=n,b=0;while(a>0){b=b*10;b=b+a%10;a=a/10;}printf("%d",b);getch();return0;}或者把后三行删掉,

#includeusingnamespacestd;doublecentigrade(doubleF){doubleC=(F-32)*5.0/9;returnC;}intmain(){doubl

#include"stdio.h"voidmain(){inta,b,i;intsum=0;printf("Inputstartyear:");scanf("%d",&a);printf("

#includeintsum(intn){ints=0;while(n){s+=n%10;n/=10;}returns;}intmain(void){ints=0;for(int

longfun(chars[]){inti;while(s[i]!='\0'){s[i]=s[i]-112;//112为字符‘0’的ASCII码,//字符和整形数可以直接运算,结果为整数输出数组用%d

#includeintmain(){intn;scanf("%d",&n);do{printf("%d",n%10);}while(n/=10);printf("\n");return0;}

voidfun(intm,intn){for(inti=m;i

swap(int&i,int&j){inttemp;temp=i;i=j;j=temp;}main(){inta,b,cscanf("%d,%d,%d",a,b,c);if(a

Reverse

#includemain(){intn;printf("Pleaseenteranumber:");scanf("%d",&n);if(n%2==0)printf("Thenumber%dis偶数\n

好吧,简单说下：voidfun(floata[][]){inti,j,floatt;for(i=0;a[i][0];i++)for(j=i;a[i][j];j++){t=a[i][j];a[i][j]

#include <stdio.h>char* dg(char* instr, char* outstr, char* 

#includeintstrl(char*s)//你需要的函数.{inti,k=0;for(i=0;s[i];i++)k++;returnk;}voidmain()//主函数,目的是测试下前面的函数.

#includevoidswap(int&a,int&b){inttem=a;a=b;b=tem;}intmain(){intm=5,n=100;printf("m=%d,n=%d\n",m,n);s

voids(folata,folatb,folatc){folatp;p=(a+b+c)/2;S=√[p(p-a)(p-b)(p-c)];returns;}

#includedoubleDistance(V&v1,V&v2){dx=v1.x-v2.x;dy=v1.y-v2.y;dz=v1.z-v2.z;returnsqrt(x*x+y*y+z*z);}其中

1:symsabRspRsdR这一句去掉就可以了.function[Rp,Rd,R]=aa11(a,b)N=[sqrt(51),sqrt(21.8);sqrt(2.3),sqrt(48.5)];N1=

//使用海伦公式#include/*ForIO*/#include/*Forsqrt()*/intmain(void){doublea,b,c,p,s;printf("请输入a,b和c：");scan

intfun(inta,intb){if((a/10)=10||(b/10)=10){std::cout

//今天被百度吞了4份代码,上帝保佑这个别被吞了,哥的分数啊.#includeintmain(){intn,i,j;scanf("%d",&n);for(i=1;i