C语言求阶乘程序无法执行
\u672c\u4ebac\u8bed\u8a00\u521d\u5b66\u8005\u6211\u60f3\u505a\u4e00\u4e2a\u9636\u4e58\u8fd0\u7b97\u7684\u7a0b\u5e8f\uff0c\u4f46\u662f\u65e0\u6cd5\u8fd0\u884c\uff0c\u6211\u627e\u4e0d\u51fa\u6765\u9519\uff0c\u671b\u5927\u795e\u5e2e\u5fd9\uff01\uff01\uff01\u628awhile(n==1) \u6539\u4e3a while(n>=1)\uff0c\u5426\u5219\u7684\u8bdd\uff0c\u5982\u679c\u4f60\u8f93\u5165\u7684\u662f5\uff0c\u90a3\u7b2c\u4e00\u6b21\u6267\u884c\u5faa\u73af\u7684\u65f6\u5019\u6761\u4ef6\u5c31\u4e0d\u6210\u7acb\uff0c\u6839\u672c\u5c31\u6ca1\u8fdb\u884c\u9636\u4e58\u7684\u8fd0\u7b97\u3002
\u8fd8\u6709\uff0c\u90a3\u4e2awhile(0) { \u8fd9\u4e24\u4e2a\u53bb\u4e86\u5427\uff0c\u753b\u86c7\u6dfb\u8db3\uff0c\u8fd8\u6dfb\u9519\u4e86\uff0cwhile\uff080\uff09\u8fd9\u6761\u4ef6\u6c38\u8fdc\u90fd\u4e0d\u4f1a\u6210\u7acb\u3002\u6539\u5b8c\u540e\u7684\u51fd\u6570\u4f53\uff1a
int n;
int count=0; //\u5176\u5b9ecount\u53d8\u91cf\u5b9a\u4e49\u7684\u6ca1\u7528\uff0c\u53ef\u4ee5\u53bb\u6389
long op=1;
printf("Please input the n for jiecheng:\n");
scanf("%d",&n);
while(n>= 1)
{
op=op * n;
n=n - 1;
count++; //\u5982\u679c\u4e0a\u9762\u7684count\u5b9a\u4e49\u53bb\u6389\uff0c\u90a3\u4e48\u8fd9\u6761\u4e5f\u53ef\u4ee5\u53bb\u6389\u4e86
}
printf("THE n!is %ld",op);
return 0; //\u4f60\u8fd9\u4e2a\u5730\u65b9\u539f\u6765\u7f3a\u5c11\u4e2a\u5206\u53f7
/*\u8be5\u7a0b\u5e8f\u8c03\u5f0f\u901a\u8fc7*/
#include
#define n 5 /*\u8be5\u503c\u81ea\u5b9a*/
main()
{
int cnt=n,i=1;
do
{
i*=cnt;
}while(--cnt);
printf("%d\n",i);
getch();
}
我觉得逻辑上应该没问题
如果输出的正确
那就是scanf有问题
因为scanf会把你输入的数字和\n当成两个数存入变量中
你这个程序不是求和的嘛，阶乘不是从1一直乘到N吗。
sacnf函数中去掉"\n"
sum = 1;
sum = sum + i 该给sum = sum * i;
你在试试。还有，最好不要用goto。
for(i = 1; i <= n; i ++)
{
sum = sum * i;
}
或者
while(i <= n)
{
sum = sum * i;
i ++;
}
#include <stdio.h>
void main()
{
int i,n,sum=0;
i=1;
printf("please input the n:");
scanf("%d",&n);
loop: if(i<=n)
{sum=sum*i;
i++;
goto loop;}
printf("%d\n",sum);
getch();
}
你这是求和啊
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