HDOJ 1216 Assistance Required

Assistance Required

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 1782 Accepted Submission(s): 943



Problem Description

After the 1997/1998 Southwestern European Regional Contest (which was held in Ulm) a large contest party took place. The organization team invented a special mode of choosing those participants that were to assist with washing the dirty dishes. The contestants
would line up in a queue, one behind the other. Each contestant got a number starting with 2 for the first one, 3 for the second one, 4 for the third one, and so on, consecutively.

The first contestant in the queue was asked for his number (which was 2). He was freed from the washing up and could party on, but every second contestant behind him had to go to the kitchen (those with numbers 4, 6, 8, etc). Then the next contestant in the
remaining queue had to tell his number. He answered 3 and was freed from assisting, but every third contestant behind him was to help (those with numbers 9, 15, 21, etc). The next in the remaining queue had number 5 and was free, but every fifth contestant
behind him was selected (those with numbers 19, 35, 49, etc). The next had number 7 and was free, but every seventh behind him had to assist, and so on.

Let us call the number of a contestant who does not need to assist with washing up a lucky number. Continuing the selection scheme, the lucky numbers are the ordered sequence 2, 3, 5, 7, 11, 13, 17, etc. Find out the lucky numbers to be prepared for the next
contest party.

Input

The input contains several test cases. Each test case consists of an integer n. You may assume that 1 <= n <= 3000. A zero follows the input for the last test case.

Output

For each test case specified by n output on a single line the n-th lucky number.

Sample Input

1
2
10
20
0

Sample Output

2
3
29
83

题意就是i从2开始 若i没被标记则每隔i个没被标记的数就标记下

模拟链表

AC CODE

#include
#include
#include
#include
using namespace std;

int main(){
list l;   //空链表
list::iterator pos = l.begin();    //
int lucky[3001];
int i(2);   //建一个含默认值是2的元素的链表
for ( ; i != 35000; ++i)
l.push_back(i);

for (i = 0; i < 3001; ++i){
lucky[i] = l.front();  //front()返回第一个元素的引用
int key = lucky[i];
pos = l.begin();   // begin()返回第一个元素的指针
while (pos != l.end()){
pos = l.erase(pos);   // 删除一个元素
int k = key - 1;
while (k--) {
if (pos == l.end())  //end()返回最后一个元素的下一位置的指针
break;
++pos;
}
}
}
int n;
while (scanf("%d", &n), n){
printf("%d\n", lucky[n - 1]);
}
return 0;
}

打表

AC CODE

#include
#define N 35000

int main(){
int arr
= {0};
int lucky
;
int i, j, k, n;
n = 0;
for (i = 2; i < N; i++){
if (arr[i] == 0){
lucky[n++] = i;
k = i + 1;
j = 0;
while (k < N){
if (arr[k] == 0)
j++;
if (j == i){
arr[k] = 1;
j = 0;
}
k++;
}
}
}
while (scanf("%d", &i), i){
printf("%d\n", lucky[i - 1]);
}
return 0;
}