怎么在c语言中写一个方程来计算一个数的n次方 c语言中一个数的n次方怎么表达
c\u8bed\u8a00\u600e\u4e48\u8868\u793a\u4e00\u4e2a\u6570\u7684n\u6b21\u65b9C\u8bed\u8a00\u4e2d\u8ba1\u7b97\u4e00\u4e2a\u6570\u7684N\u6b21\u65b9\u53ef\u4ee5\u7528\u5e93\u51fd\u6570pow\u6765\u5b9e\u73b0\u3002\u51fd\u6570\u539f\u578b\uff1adouble pow(double x, double y)\u3002
\u4ee3\u7801\u5982\u4e0b\uff1a
#include
#include
int main( )
{
printf("%f",pow\uff08x\uff0cy)\uff09\uff1b
return 0\uff1b
}
\u6ce8\uff1a\u4f7f\u7528pow\u51fd\u6570\u65f6\uff0c\u9700\u8981\u5c06\u5934\u6587\u4ef6#include\u5305\u542b\u8fdb\u6e90\u6587\u4ef6\u4e2d\u3002\u3001
\u6269\u5c55\u8d44\u6599\uff1a
\u5176\u4ed6\u65b9\u6cd5\u8868\u793a\u4e00\u4e2a\u6570\u7684n\u6b21\u65b9\uff1a
#include
int main( )
{ int i,k = n\uff1b for(i = 1\uff1bi < n\uff1bi++)
{ k *= 2\uff1b
}
printf("%d",k)\uff1b
return 0\uff1b
}
^
\u5728c\u8bed\u8a00\u4e0d\u662f\u6c42\u65b9
\u5728vb\u5c31\u662f.
\u6240\u4ee5
a^n
\u8fd9\u6837\u662f\u4e0d\u884c\u7684
\u6240\u4ee5c\u91cc\u9762\u53ea\u80fd\u7528
pow(x,y)
\u5982\u679c\u5728c++\u91cc\u9762\u5012\u662f\u53ef\u4ee5\u91cd\u8f7d
^
,\u7136\u540e\u5c31\u53ef\u4ee5a^n\u4e86,\u5728c\u91cc\u9762\u4e00\u4e2a\u662f\u4e0d\u80fd\u8fd9\u6837.\u56e0\u4e3a\u4e0d\u652f\u6301\u64cd\u4f5c\u7b26\u91cd\u8f7d
思路：定义一个函数fun(x,n)求x的n次方，即进行n次for循环执行x的累成，主函数调用fun函数。
参考代码：
#include <stdio.h>
int fun(int x,int n){
int s=1;
while(n--){
s*=x;
}
return s;
}
int main()
{
int x=2,y=10;
printf("%d
",fun(2,10));
return 0;
}
/*
运行结果：求2的10次方
1024
*/
#include<stdio.h>
int main(void)
{
int n,i;//n表示一个整数，i表示幂
int sum=1;
printf("请输入一个数和幂:\n");
scanf("%d%d",&n,&i);
for(int j=0;j<i;j++)
{
sum=sum*n;
}
printf("%d\n",sum);
return 0;
}
#include <stdio.h>
double powern( double n,double x )
{
double d=1;
int i;
for( i=0;i<n;i++ )
d *= x ;
return d ;
}
void main()
{
double xn,x,n;
x=2;
n=10;
xn=powern(n,x);
printf("%g^%g=%g
", x, n, xn );
}
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绛旓細include<stdio.h>#include<math.h>main(){int a,b,c,d,x1,x2;printf("璇蜂緷娆¤緭鍏锛宐锛c鐨鍊硷細");scanf("%d %d %d",&a,&b,&c); d=b*b-4*a*c;if(d<0) printf("姝鏂圭▼鏃犲疄鏍癸紒");else{x1=(-b+sqrt(d))/(2*a);x2=(-b-sqrt(d))/(2*a);printf("x1 = %d ...
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绛旓細} void n(float a,float b,float c){ double x;x=(-b)/(2*a);printf("鏂圭▼鐨鏍逛负%.2lf",x);} void f(float a,float b,float c){ printf("鏂圭▼鏃犲疄鏁版牴\n");} main(){ float a,b,c;printf("璇疯緭鍏,b,c鐨鍊糪n");scanf("%f%f%f",&a,&b,&c);if(b*b-4*a*c>...
