P1145 约瑟夫
思路: 用链表进行模拟。
1.不用STL
#include <iostream>
using namespace std;
struct Node{
int n;
Node *next;
};
long n,m;
Node *head,*p,*r;
int main(){
cin >> n >> m;
head = new Node;
head->n = 1;
head->next = NULL;
r = head;
for(int i=2;i<=n;++i){
p = new Node;
p->n = i;
p->next = NULL;
r->next = p;
r = p;
}
r->next = head;
r = head;
if(m != 1){
for(int i=0;i<n;++i){
for(int j=0;j<m-2;++j){
r = r->next;
}
cout << r->next->n << " ";
p = r->next;
r->next = r->next->next;
r = r->next;
delete p;
}
}else{
for(int i=0;i<n;++i){
cout << r->n << " ";
p = r;
r = r->next;
delete p;
}
}
cout << endl;
return 0;
}
2. 使用STL
#include <iostream>
#include <list>
using namespace std;
long n,m;
list<int> l;
int main() {
cin >> n >> m;
if(n<0 && m<0) {
return 0;
}
for(int i=1; i<=n; ++i) {
l.push_back(i);
}
list<int>::iterator beg = l.begin();
for(int i=0; i<n; ++i) {
for(int j=1; j<m; ++j) {
beg++;
if(beg == l.end()) {
beg = l.begin();//因为List内容不断变化,所以应该重新获取一次R。
}
}
cout << *beg << " ";
beg = l.erase(beg);
if(beg == l.end()) {
beg = l.begin();//因为List内容不断变化,所以应该重新获取一次R。
}
}
cout << endl;
return 0;
}