c语言X的N次幂
c\u8bed\u8a00x\u7684n\u6b21\u5e42double x,p; //\u5b9a\u4e49\u4e24\u4e2a\u53cc\u7cbe\u5ea6\u6d6e\u70b9\u6570\uff0cx\u548cp\uff0c\u540c\u65f6\u5206\u914d\u5b58\u50a8\u5355\u5143
unsigned n; //\u5b9a\u4e49\u4e00\u4e2a\u65e0\u7b26\u53f7\u6574\u6570\uff0cn\uff0c\u540c\u65f6\u5206\u914d\u5b58\u50a8\u5355\u5143
printf("Please input x:"); //\u8f93\u51fa"Please input x:"\uff0c\u63d0\u793a\u7528\u6237\u8f93\u5165x\u7684\u503c
scanf("%lf",&x); //\u8f93\u5165\u4e00\u4e2a\u6d6e\u70b9\u6570\uff0c\u5b58\u5728&x\u6240\u6307\u5411\u7684\u5b58\u50a8\u5355\u5143
printf("Please input n:"); //\u8f93\u51fa"Please input n:"\uff0c\u63d0\u793a\u7528\u6237\u8f93\u5165n\u7684\u503c
scanf("%d",&n); //\u8f93\u5165\u4e00\u4e2a\u6574\u6570\uff0c\u5b58\u5728&n\u6240\u6307\u5411\u7684\u5b58\u50a8\u5355\u5143
p=1.0; //\u7ed9p\u8d4b\u503c\u4e3a1.0
while(n--) p*=x; \u2026\u2026\u2026\u2026#//\u8fd9\u662f\u4e00\u4e2a\u5faa\u73af\uff0c\u5728while\u540e\u7684\u62ec\u53f7\u5185\u5185\u5bb9\u4e3a\u771f\u65f6\uff0c\u8fdb\u5165\u5faa\u73af
\u8bed\u53e5\u4e3an--\uff0c\u5c31\u662f\u5148\u5224\u65adn\u7684\u503c\u662f\u5426\u4e3a\u771f\uff0c\u518d\u6267\u884cn=n-1\u64cd\u4f5c
\u82e5n\u4e0d\u7b49\u4e8e0\uff0c\u5373\u8bed\u53e5\u4e3a\u771f
\u8bed\u53e5\u4e3a\u771f\u8fdb\u5165p*=x\u7684\u8bed\u53e5\u5faa\u73af
\u5373p=p*x,\u5176\u4e2d\uff0cp\u7684\u521d\u59cb\u503c\u4e3a1.0\uff0cx\u662f\u4f60\u8f93\u5165\u7684\u67d0\u6d6e\u70b9\u6570
\u5f53n\u4e3a\u96f6\u65f6\u9000\u51fa\u5faa\u73af\uff0c\u6b64\u65f6\uff0c\u4e00\u5171\u6267\u884c\u4e86n\u6b21*x\u7684\u64cd\u4f5c\uff0c\u5373x\u7684n\u6b21\u65b9
p\u4e2d\u5b58\u653e\u7684\u662fx\u7684n\u6b21\u65b9\u7684\u7ed3\u679c
printf("%f",p); //\u8f93\u51fap\u7684\u503c
long mi_func(long a, int n){ int i =1; long ret = a; if ((0 == a) && (0 == n)) { printf("Invalid parameter!\n"); } else if(0 == a) { return 0; } else if (0 == n) { return 1; } for(i = 1; i <= n; i++) ret*= ret; return ret;}
输出是不是有问题啊?应该这样吧:
输入
2 (repeat=2)
1.5 2
输出
2.25
输入
2.0 10
输出
1024.00
如果这样可以这样写:
#include <stdio.h>
int main( )
{
int ri, repeat;
int i, n;
double x, mypow;
scanf("%d", &repeat);
for(ri=1; ri<=repeat; ri++)
{
scanf("%lf%d", &x, &n);
for(mypow=1.0, i=0; i<n; i++)
mypow*=x;
printf("%.2f\n", mypow);
}
return 0;
}
值得注意的是:编写风格不对,int main()是ANSI C++标准,怎么后面程序用的是C语言库函数写的? 而且没有RETURN语句呼应!
mypow=1;
for (i=1;i<=n;i++)
mypow*=x;
for(i=1,mypow=1;i<=n;i++) mypow*=x;
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绛旓細{ if
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绛旓細include <stdio.h> include <math.h> int main(){ double x;int n;scanf("%lf %d",&x,&n);printf("%.1lf\n",pow(x,n));return 0;}
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