单选题：以下程序段（ ）的功能是计算序列 1 + 1/2 + 1/3 + ... 的前N项之和。

以下程序段（ ）的功能是计算序列 1 + 1/2 + 1/3 + ... 的前N项之和。 @[C](1)
A.
```
int i, n, sum;
scanf("%d", &n);
sum = 0;
for (i = 1; i <= n; i++){
sum = sum + 1.0/i;
}
```
B.
```
int i, n;
double sum;
scanf("%d", &n);
for (i = 1; i <= n; i++){
sum = sum + 1.0/i;
}
```
C.
```
int i, n;
double sum;
scanf("%d", &n);
sum = 0;
for (i = 1; i <= n; i++){
sum = sum + 1.0/i;
}
```
D.
```
int i, n;
double sum;
scanf("%d", &n);
sum = 0;
for (i = 1; i <= n; i++){
sum = sum + 1/i;
}
```
E.
```
int i, n;
double sum;
scanf("%d", &n);
sum = 0;
for (i = 1, i <= n, i++){
sum = sum + 1.0/i;
}
```





A.
```
int i, n, sum;
scanf("%d", &n);
sum = 0;
for (i = 1; i <= n; i++){
sum = sum + 1.0/i;
}
```
B.
```
int i, n;
double sum;
scanf("%d", &n);
for (i = 1; i <= n; i++){
sum = sum + 1.0/i;
}
```
C.
```
int i, n;
double sum;
scanf("%d", &n);
sum = 0;
for (i = 1; i <= n; i++){
sum = sum + 1.0/i;
}
```
D.
```
int i, n;
double sum;
scanf("%d", &n);
sum = 0;
for (i = 1; i <= n; i++){
sum = sum + 1/i;
}
```
E.
```
int i, n;
double sum;
scanf("%d", &n);
sum = 0;
for (i = 1, i <= n, i++){
sum = sum + 1.0/i;
}
```


答案：C