PAT基础编程题目-6-6 求单链表结点的阶乘和
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2022-03-13 20:47:53
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PAT基础编程题目-6-6 求单链表结点的阶乘和
题目详情
解答
C语言版
#include <stdio.h>
#include <stdlib.h>
typedef struct Node* PtrToNode;
struct Node
int Data; // 存储结点数据
PtrToNode Next; // 指向下一个结点的指针
};
typedef PtrToNode List; // 定义单链表类型
int FactorialSum(List L);
int main()
{
int N, i;
List L, p;
scanf("%d", &N);
L = NULL;
for (i = 0; i < N; i++) {
p = (List)malloc(sizeof(struct Node));
if (p) {
scanf_s("%d", &p->Data);
p->Next = L;
L = p;
}
}
printf("%d\n", FactorialSum(L));
return 0;
}
int FactorialSum(List L)
{
int sum =0, product;
while(L) {
product = 1; //乘积
for (int i = 2; i <= L->Data; i++)
{
product = product * i;
}
sum = sum + product;
L = L->Next;
}
return sum;
}
C++版
#include<iostream>
using namespace std;
typedef struct Node {
int Data;
struct Node* Next;
}Node, * List;
int FactorialSum(List L);
int main() {
int n;
cin >> n;
List L=NULL, p;
while (n--) {
p = (Node*)malloc(sizeof(Node));
if (p) {
cin >> p->Data;
p->Next = L;
L = p;
}
}
cout << FactorialSum(L);
return 0;
}
int FactorialSum(List L)
{
int sum = 0, product;
while (L) {
product = 1; //乘积
for (int i = 2; i <= L->Data; i++)
{
product = product * i;
}
sum = sum + product;
L = L->Next;
}
return sum;
}
Java版
public class Main{
private static class Node {
int Data;
Node Next;
}
private static int FactorialSum(Node L) {
int sum = 0;
int product;
while(L!=null) {
product = 1;
for (int i = 2; i <= L.Data; i++) {
product = product * i;
}
sum = sum +product;
L = L.Next;
}
return sum;
}
public static void main(String[] args) {
int n;
Node L= null;
Scanner scanner = new Scanner(System.in);
if(scanner.hasNext()) {
n = scanner.nextInt();
for (int i = 0; i < n; i++) {
Node p = new Node();
p.Data = scanner.nextInt();
p.Next = L;
L = p;
}
}
scanner.close();
System.out.println(FactorialSum(L));
}
}
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