如题所述
#include <stdio.h>
int f1(int n){
if (n == 1) {
return 1;
}
return n*f1(n-1);
}
int main() {
int m,n, sum;
scanf("%d,%d", &m,&n);
sum = f1(m) + f1(n);
printf(" result = %d\n",sum);
return 0;
}
int f1(int n){
if (n == 1) {
return 1;
}
return n*f1(n-1);
}
int main() {
int m,n, sum;
scanf("%d,%d", &m,&n);
sum = f1(m) + f1(n);
printf(" result = %d\n",sum);
return 0;
}
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第1个回答 2015-06-02
//参考:
#include <stdio.h>
int jc(int n)
{
int i, ans = 1;
for (i = 1; i <= n; ++i){
ans *= i;
}
return ans;
}
int main()
{
int m, n;
scanf ("%d%d", &m, &n);
int ans = jc(m)+jc(n);
printf ("%d\n", ans);
return 0;
}追问
#include <stdio.h>
int jc(int n)
{
int i, ans = 1;
for (i = 1; i <= n; ++i){
ans *= i;
}
return ans;
}
int main()
{
int m, n;
scanf ("%d%d", &m, &n);
int ans = jc(m)+jc(n);
printf ("%d\n", ans);
return 0;
}追问
int ans = jc(m)+jc(n);这个有问题,这是什么意思
追答调用函数计算m!+n!
jc(m)计算m!
jc(n)计算n!
第2个回答 2018-05-17
#include<stdio.h>
main()
{
int n,m,value;
printf("please input n,m:\n");
scanf("%d %d",&n,&m);
value=factor1(n)+factor1(m);
printf("%d的阶乘加%d的阶乘是%d\n",n,m,value);
}
int factor1(int n,int m)
{
int result1;
int result2;
if(n==1){
return 1;
}else{
result1=factor1(n-1)*n;
return result1;}
if(m==1){
return 1;
}else{
result2=factor1(m-1)*m;
return result2;}
}
main()
{
int n,m,value;
printf("please input n,m:\n");
scanf("%d %d",&n,&m);
value=factor1(n)+factor1(m);
printf("%d的阶乘加%d的阶乘是%d\n",n,m,value);
}
int factor1(int n,int m)
{
int result1;
int result2;
if(n==1){
return 1;
}else{
result1=factor1(n-1)*n;
return result1;}
if(m==1){
return 1;
}else{
result2=factor1(m-1)*m;
return result2;}
}