本文共 1250 字,大约阅读时间需要 4 分钟。
分用三次约瑟夫环数学方法就行了。具体见
//#pragma comment(linker, "/STACK:1024000000,1024000000")#include #include #include #include #include #include #include #include #include #include #include #include #include #include using namespace std;typedef long long ll;typedef pair pii;#define pb(a) push_back(a)#define INF 0x1f1f1f1f#define lson idx<<1,l,mid#define rson idx<<1|1,mid+1,r#define PI 3.1415926535898template T min(const T& a,const T& b,const T& c){ return min(min(a,b),min(a,c));}template T max(const T& a,const T& b,const T& c){ return max(max(a,b),max(a,c));}void debug(){#ifdef ONLINE_JUDGE#else freopen("d:\\in.txt","r",stdin); freopen("d:\\out1.txt","w",stdout);#endif}char getch(){ char ch; while((ch=getchar())!=EOF) { if(ch!=' '&&ch!='\n')return ch; } return EOF;}int main(){ int t; scanf("%d",&t); for(int ca=1;ca<=t;ca++){ int n,k; scanf("%d%d",&n,&k); int x; x=(k+2)%3; for(int i=4;i<=n;i++) x=(x+k)%i; printf("%d ",x+1); x=(k+1)%2; for(int i=3;i<=n;i++) x=(x+k)%i; printf("%d ",x+1); x=0; for(int i=2;i<=n;i++) x=(x+k)%i; printf("%d\n",x+1); } return 0;}
转载于:https://www.cnblogs.com/BMan/p/3264283.html