编写一个函数,获取一个整型值,并将数字反向.例如数值为789,函数将返回987.-搜答案-搜搜考试网

没必要用循环,matlab有求最大值的函数：function[rowindex,colindex]=shiyan1(a)[colmaxrows]=max(a);[maxvaluecol]=max(co

#defineN10#include"stdio.h"#include"math.h"intifp(inta[]);voidmain(){inti;inta[N];for(i=0;i再问：��C��

#includelongfun(longx){intn=10,m=0;while(x){intt=x%10;if(t%2==0){m+=t*n/10;n*=10;}x/=10;}returnm;}vo

a=12a*a=144a-=a*a,此时a=-132,同时也是这个表达式的值；a+=a-=a*a,此时a=-264,同时也是这个表达式的值.代码如下：#includemain(){inta=12,b;

/*90897:7980923098:8903233201:1023378653:3568736451:1546383409:90438Pressanykeytocontinue*/#include&

用sizeof(变量类型）得到字节数sizeof(longint)sizeof(int)sizeof(short)sizeof(unsignedint)字节数*8得到总位数有符号数,符号要占去1位,其

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

#includeusingnamespacestd;doubles(inta[],intn){longsum=0;int*pa=a;for(inti=0;i

#includeusingnamespacestd;voidfun2(inta[],intn,int&c1,int&c2){for(inti=0;i

这道题目也不是很难,自家先做一下.思路很清晰,没有涉及到算法问题.

voidmain(){\x05inta,b,temp,m;\x05\x05scanf("%d%d",&a,&b);if(a

#include#includevoidmaxp(int**ia,intr,intc,int*x,int*y,int*max){*max=ia[0][0];inti,j;for(i=0;i

#includeintmax(inti,intj){return(i>j)?i:j;}intmin(inti,intj){return(i

Function isum(n As Integer) As Double isum&nbs

#include#includemain(){\x09longintx;\x09inti=0;printf("input:");\x09scanf("%ld",&x);do{x=x/10;i++;\x

#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

#includemain(){inta[20],x,i;printf("输入20个数字:");for(i=0;i

好长啊,自己编吧==!

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=

voidAddTwo(int&a){for(inti=0;i