c语言X的立方 用c语言迭代法实现x=a的立方
x\u76843\u6b21\u65b9\u7528C\u8bed\u8a00\u600e\u4e48\u8868\u793a\u53ef\u4ee5\u4f7f\u7528pow(x\uff0cy)\u51fd\u6570\u6765\u5b9e\u73b0x\u76843\u6b21\u65b9\uff1b
\u8be5\u51fd\u6570\u9700\u8981\u8c03\u7528math.h\u5934\u6587\u4ef6
\u4ee3\u7801\u4e3e\u4f8b\u5982\u4e0b\uff1ax=3\u65f6
#include
#include
int main()
{
int x=3,y;
y=pow(x,3);
printf("%d\n",y);
}
\u8fd0\u884c\u7ed3\u679c\u5982\u4e0b\u56fe\uff1a
\u62d3\u5c55\u8d44\u6599\u8bed\u8a00\u662f\u4e00\u95e8\u901a\u7528\u8ba1\u7b97\u673a\u7f16\u7a0b\u8bed\u8a00\uff0c\u5e94\u7528\u5e7f\u6cdb\u3002C\u8bed\u8a00\u7684\u8bbe\u8ba1\u76ee\u6807\u662f\u63d0\u4f9b\u4e00\u79cd\u80fd\u4ee5\u7b80\u6613\u7684\u65b9\u5f0f\u7f16\u8bd1\u3001\u5904\u7406\u4f4e\u7ea7\u5b58\u50a8\u5668\u3001\u4ea7\u751f\u5c11\u91cf\u7684\u673a\u5668\u7801\u4ee5\u53ca\u4e0d\u9700\u8981\u4efb\u4f55\u8fd0\u884c\u73af\u5883\u652f\u6301\u4fbf\u80fd\u8fd0\u884c\u7684\u7f16\u7a0b\u8bed\u8a00\u3002
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import java.util.*;
public class T {
void show(int x){
System.out.println("\u8f93\u51fa\u7acb\u65b9\uff1a"+Math.pow(x, 3));
}
public static void main(String[] args){
T t=new T();
Scanner sc=new Scanner(System.in);
System.out.println("\u8bf7\u8f93\u5165\u4f60\u81ea\u5b9a\u4e49\u7684\u6570x:");
int x=sc.nextInt();
t.show(x);
}
}
\u8fd0\u884c\u7ed3\u679c\u5982\u4e0b\uff1a
\u8bf7\u8f93\u5165\u4f60\u81ea\u5b9a\u4e49\u7684\u6570x:
3
\u8f93\u51fa\u7acb\u65b9\uff1a27.0
\u8fd9\u662f\u6211\u7528java\u4ee3\u7801\u7ed9\u4f60\u5199\u7684\uff0c\u4e0d\u8fc7\u6ce8\u610f\uff1a\u6211\u5e76\u6ca1\u6709\u7528\u8fed\u4ee3\u6cd5\uff0c\u7b49\u4e00\u4f1a\u6211\u7ed9\u4f60\u53d1\u8fc7\u8fed\u4ee3\u6cd5\u7684\u505a\u6cd5\uff1b
int power(int x,int n) ; //函数声明
void main()
{
int x; //底数x
int n; //次方n
int result；//结果
printf("请分别输入底数，次方:\n");
scanf("d%d%",&x，&n)； //从键盘读取底数和次方
result=power(x,n); //调用外部算次方函数
printf（"%d的%d次方是：%d"，&x，&n,&result）;
}
int power(int x,int n)
{
int i;
int pow=1;
for(i=1;i<=n;i++) //用for 循环来求X的N次方
{
power*=x;
}
return power; //结果返回
}
我直接在这上面写的，没有在编译器上调试，看懂了知道思路就可以了
已经是最详细的了
#include<stdio.h>
int cube(int x)
{
int cube;
for(int i=0;i<3;i++)
{
cube*=x;
}
}
void main()
{
int x;
scanf("%d",&x);
printf("\n%d",cube(x));
}
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绛旓細c璇█骞虫柟鐨勮〃绀烘柟娉曟槸锛氣*鈥濄傝繖鎰忓懗鐫锛屽鏋滄垜浠兂瑕佽绠椾竴涓暟鐨勫钩鏂癸紝鎴戜滑鍙互灏嗚繖涓暟涓庡畠鑷韩鐩镐箻銆備緥濡傦紝濡傛灉鎴戜滑鎯宠璁＄畻鏁板瓧5鐨勫钩鏂癸紝鎴戜滑鍙互鍐欐垚5*5锛岀粨鏋滃皢鏄25銆傝繖鏄洜涓轰箻娉曡繍绠楃*鍏锋湁灏嗕袱涓暟鐩镐箻鐨勫姛鑳姐傚綋鎴戜滑浣跨敤涔樻硶杩愮畻绗*鏉ヨ绠椾竴涓暟鐨勫钩鏂规椂锛屾垜浠皢璇ユ暟浣滀负涔樻硶杩愮畻绗︾殑绗...
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绛旓細pingfang(x) - pingfang(y);}int lifangcha(int x,int y){return lifang(x) - lifang(y);}int main (void){int a=5,b=3;printf("a骞虫柟[%d] b骞虫柟[%d]\n",pingfang(a),pingfang(b));printf("a绔嬫柟[%d] b绔嬫柟[%d]\n",lifang(a),lifang(b));printf("a,b骞虫柟宸甗...
