单选题:以下程序段( )的功能是计算n的阶乘,假设计算结果不超过双精度范围。
以下程序段( )的功能是计算n的阶乘,假设计算结果不超过双精度范围。 @[D](1)
A.
```
int i, n;
double product;
scanf("%d", &n);
product = 0;
for (i = 1; i <= n; i++){
product = product * i;
}
```
B.
```
int i, n, product;
scanf("%d", &n);
product = 1;
for (i = 1; i <= n; i++){
product = product * i;
}
```
C.
```
int i, n;
double product;
scanf("%d", &n);
for (i = 1; i <= n; i++){
product = product * i;
}
```
D.
```
int i, n;
double product;
scanf("%d", &n);
product = 1;
for (i = 1; i <= n; i++){
product = product * i;
}
```
A.
```
int i, n;
double product;
scanf("%d", &n);
product = 0;
for (i = 1; i <= n; i++){
product = product * i;
}
```
B.
```
int i, n, product;
scanf("%d", &n);
product = 1;
for (i = 1; i <= n; i++){
product = product * i;
}
```
C.
```
int i, n;
double product;
scanf("%d", &n);
for (i = 1; i <= n; i++){
product = product * i;
}
```
D.
```
int i, n;
double product;
scanf("%d", &n);
product = 1;
for (i = 1; i <= n; i++){
product = product * i;
}
```
答案:D
A.
```
int i, n;
double product;
scanf("%d", &n);
product = 0;
for (i = 1; i <= n; i++){
product = product * i;
}
```
B.
```
int i, n, product;
scanf("%d", &n);
product = 1;
for (i = 1; i <= n; i++){
product = product * i;
}
```
C.
```
int i, n;
double product;
scanf("%d", &n);
for (i = 1; i <= n; i++){
product = product * i;
}
```
D.
```
int i, n;
double product;
scanf("%d", &n);
product = 1;
for (i = 1; i <= n; i++){
product = product * i;
}
```
A.
```
int i, n;
double product;
scanf("%d", &n);
product = 0;
for (i = 1; i <= n; i++){
product = product * i;
}
```
B.
```
int i, n, product;
scanf("%d", &n);
product = 1;
for (i = 1; i <= n; i++){
product = product * i;
}
```
C.
```
int i, n;
double product;
scanf("%d", &n);
for (i = 1; i <= n; i++){
product = product * i;
}
```
D.
```
int i, n;
double product;
scanf("%d", &n);
product = 1;
for (i = 1; i <= n; i++){
product = product * i;
}
```
答案:D