一个简单编程题#include#includeint main(){int n,i=1,m=1;double x,s=0
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一个简单编程题
#include
#include
int main(){
int n,i=1,m=1;double x,s=0;
scanf("%d%d",&x,&n);
while(i
#include
#include
int main(){
int n,i=1,m=1;double x,s=0;
scanf("%d%d",&x,&n);
while(i
我帮你编两个函数吧,调用请自己琢磨下.
double fact(int n)
{
double m = n;
if(n < 2) return 1;
return m * fact(n - 1);
}
double mypow(double x, int n)
{
double y = 1.0;
for (int i = 0; i < n; i++) y *= x;
return y;
}
提示:循环时可用:
s += mypow(-1, i-1) * mypow(s, i) / fact(i);
double fact(int n)
{
double m = n;
if(n < 2) return 1;
return m * fact(n - 1);
}
double mypow(double x, int n)
{
double y = 1.0;
for (int i = 0; i < n; i++) y *= x;
return y;
}
提示:循环时可用:
s += mypow(-1, i-1) * mypow(s, i) / fact(i);
#includeint main(void){int x,y,t,n,flag=1,s,i;while( scanf("
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