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北京大学《ACM-ICPC竞赛训练》暑期课面向全球招生。容量有限,报名从速!

源码及我错过的点,希望能有帮助。。

Posted by 1600017830 at 2017-01-12 13:22:00 on Problem 1012
1.不是真的采用循环链表模拟,采用数学方法,利用取余来模拟踢出的过程;我们关心的只是好人还是坏人,原来的序号不重要,只要取余的余数小于k就是取到好人,break。
2.递增时要加k+1,而不是+1;
3.存数据,不存会超时。
贴源码:

#include<iostream>
using namespace std;
int finalans[30] = { 0 };//存储已经计算好的结果
int main()
{
int n = 0, m = 0, i = 0, judge = 1, judge2 = 1;
int ans = 0;
while (cin >> n&&n!=0)
{
if (finalans[n] == 0)
{
for (m = n;;m += n + 1)//尽可能快的递增
{
{
for (i = 0;i < n;i++)
{
ans = (m + ans) % (2 * n - i);//计算余数
if (ans < n)
{
judge = 0;
break;
}
}
ans = 0;
for (i = 0;i < n;i++)
{
ans = (m + 1 + ans) % (2 * n - i);
if (ans < n)
{
judge2 = 0;
break;
}
}
if (judge == 0 && judge2 == 0)
{
judge = 1;
judge2 = 1;
ans = 0;
continue;
}
else
{
if (judge == 1)
{
cout << m + 1 << endl;
finalans[n] = m + 1;
}
else if (judge2 == 1)
{
cout << m + 2 << endl;
finalans[n] = m + 2;
}
ans = 0;
break;
}
}
}
}
else
cout << finalans[n] << endl;
}
system("pause");
return 0;
}

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