c语言函数调用编写程序求1*1*1+2*2*2+3*3*3+4*4*4+5*5*5 C语言 用函数调用实现 1!+2!+3!+4!+5!+6!
c\u8bed\u8a00\u51fd\u6570\u8c03\u7528\u7f16\u5199\u7a0b\u5e8f\u6c421*1*1+2*2*2+3*3*3+4*4*4+5*5*5#includeint fun(int n){return n*n*n;}int main(){int i=1,s=0; for(i=1;i<6;i++) s+=fun(i); printf("%d\n",s); return 0;}
#includeint jiecheng(int n){int t = 1, i;for (i = 1; i <= n; i++){t = t*i;}return t;}int main(void){int i, sum = 0;int n = 6;for (i = 1; i <= n; i++)sum = sum + jiecheng(i);printf("1!+2!+3!+4!+5!+6!=%d\n", sum);return 0;}
自定义一个int f(int n)函数递归调用来实现。代码如下：
#include "stdio.h"
int f(int n){
if(n==1)
return 1;//终了返回1
return n*n*n+f(n-1);//n^3+递归调用f求和
}
int main(int argc,char *argv[]){
printf("1*1*1+2*2*2+3*3*3+4*4*4+5*5*5 = %d
",f(5));//用5调用f函数并输出
return 0;
}
运行结果：
#include <stdio.h>
int f1(int ti)
{
return ti * ti * ti;
}
int f2(int &s, int i)
{
return s += f1(i);
}
void main()
{
int sum = 0,i = 0;
for(int i = 1; i <= 5; ++i)
f2(sum,i);
printf("%d\n",sum);
getchar();
}
f1(int N){ int i=1; long num=0; for(i,i<=N,i++) { num=num+f1(i) }} f1(int n){ long xnum; xnum=n*n*n; return xnum;}
绛旓細include "stdio.h"int f(int n){if(n==1)return 1;//缁堜簡杩斿洖1return n*n*n+f(n-1);//n^3+閫掑綊璋冪敤f姹傚拰}int main(int argc,char *argv[]){printf("1*1*1+2*2*2+3*3*3+4*4*4+5*5*5 = %d\n",f(5));//鐢5璋冪敤f鍑芥暟骞惰緭鍑簉eturn 0;}杩愯缁撴灉锛...
绛旓細include<stdio.h>int fun(int n){return n*n*n;}int main(){int i=1,s=0; for(i=1;i<6;i++) s+=fun(i); printf("%d\n",s); return 0;}
绛旓細} int fun(n) ///閫掑綊璋冪敤 { if(n==1)return 1;else return n*n+fun(n-1);}
绛旓細include<stdio.h>void main(){ int i,j; for(i=1;i<10;i++) { for(j=1;j<=i;j++) printf("%d*%d=%2d ",i,j,i*j); printf("\n"); }}
绛旓細include<stdio.h> int main(){ int i,sum;for(i=1,sum=0;i<=100;i++){ sum=i*i+sum;} printf("%d\n",sum);return 0;}
绛旓細"璇疯緭鍏鐨勫:\n"); scanf("%d", &n); double s = 0; for (int i = 1; i <= n; i++) { //杩欓噷瑕佹敞鎰忎笅锛岄渶瑕佽浆鎴愭诞鐐瑰瀷鐩搁櫎锛屽惁鍒欑粨鏋滄案杩滄槸0 s = 1.0/calc(n); } printf("s = %lf\n", s);}杩愯缁撴灉锛...
绛旓細include<stdio.h> int f(int i){ if(i==1)return 1;return f(i-1)*i;} int main(){ int i;for(i=1;i<=5;i++)printf("%d\n",f(i));return 0;}
绛旓細long int Sum(void){ long int sum_n=0;for(n=1;n<98;n++){ sum_n +=n*(n+1)*(n+2);} return sum_n;}
绛旓細double diao(int k){ double l=1;int j=1;for(;j<=k;j++)l*=j;return l;} main(){ int q=1,i=1;for( ;q<=20;q++){ if(i%7==0)printf("\n");printf(" %.0lf ",diao(q));i++;} } 涓嶇煡閬撲綘鏄笉鏄姹1-20涓暟鍚勮嚜鐨勯樁灞,濡傛灉鏄叏閮ㄧ殑,浣犵敤涓涓鍙ュ皢瀹冧滑鍏...
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