c 编写一函数,求任意两个m*n矩阵的和和差
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/03 07:04:10
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,
#includelongfactorial(intm,intn){longsum=1,sum1=1;inti;if(m-n>n){for(i=m;i>m-n;i--)sum*=i;for
楼上两位的代码可能有点小问题s += m*m+1/m;改成s += m*m+1.0/m;再问:您好,请您帮忙写一个完整的编程好么?谢谢了再答:#include&
#include#include#include/*利用辗转相除法求最大公约数*/intgcd(intn,intm){intr;if(n
fun(intn){intc=0;whlie(n>0){c=n%10;printf("%d",c);n=n/10;}}
想了想...1--varn,m:integer;functionf(n,m:integer):longint;vari,s:longint;{因为总和可能很大所以用longint}beginfori:
intjc(intx){returnx==1?1:jc(x-1)*x;}再问:可以编一个完整的么?我直接运行试一下。。。新手,不好意思,,,,谢谢再答:intjc(intx){returnx==1?1
intfib(n){if(n
//fibonacci数列:1123581321...#include#includeintmain(void){longa=1;longb=1;intn;intk;printf("inputnumb
#include#include#defineX3#defineY3inta[X][Y];intb[X][Y];intc[X][Y];voidmatrix(intb[][X],intc[][Y]);v
以下代码基本能够符合您的要求:#includeintfac(intn){intm=0;if(n==1){return1;}else{returnn*fac(n-1);}}doubleA(intn,in
#includevoidmain(){inta,b,num1,num2,temp;scanf("%d%d",&num1,&num2);if(num1
%编成M函数文件运行后,在命令窗口输入要知道的自然数n,即可求得对应项的Fibonacci数列%有哪步有疑问请问user_entry=input('Pleaseenterthenumberyouwan
第一题:#includevoidmain(){inta[10]={1,2,3,4,5,6,7,8,9,10},i,max,min;/*初始化的值任意定,只要是在整型范围内都行*/max=a[0];
#includeintcal(intm,intn){intret=0;ret=m%n;returnret;}intmain(intargc,char**argv){intm,n,max,min
占天时地利人和取九州四海财宝横批:财源不断
#include <stdio.h>int abc(int x,int y);void main(){int n1,n2,i;
#includeusingnamespacestd;voidmain(){inta=0,b=0;cin>>a>>b;cout
#includevoidmain(){inta,b,sum=0;printf("请输入两个整数:");//将两个改成n个就好了scanf("%d%d",&a,&b);sum=a+b;printf("%
functionf=d(n)f(1)=1;f(2)=1;fori=3:nf(i)=f(i-1)+f(i-2);end