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自己推了一个公式 ：）

Posted by 888899977 at 2007-10-12 17:25:56 on Problem 1012 
int f(int n,int m,int t)
{
    if(t==1) return (m-1)%n;
    else return (m%n+f(n-1,m,t-1))%n;
}//n人报m第t轮出列的人的编号(从0到n-1)

这样就可以ac了。
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