HDU 1027 Ignatius and the Princess II (模拟题)

题意:   给出n,m求第m小序列, (1.2.3.4....n)     (  1<=n<=1000, 1<=m<=10000   )

解法:   根据m的数据大小,  至多改变后面8位,  也就是8!,   可以进行模拟,   确定需要改变后面k位，然后,先输出前n-k位数, 对第n-k+1到n位的数进行模拟.

code:

#include
#include
#include
#include
#include
#include
#include
#define inf 0x3f3f3f3f3f
using namespace std;
const int maxn = 100;
int num[11];
int sum[11];
void init(){
	int i=1; 
	sum[0]=1;
	while(i<10){
		sum[i] = sum[i-1]*i;
		i++;
	}
	return ;
}
int mark[11];
int main(){
   //freopen("in.txt", "r", stdin);
   init();
   int n,m ,i, j, k;
   sum[0]=1;
   while(~scanf("%d %d", &n, &m)){
   	    for(i=1; i<=9; i++){
   	    	if(sum[i] >= m) break;
		}
		for(j=1; j<=n-i; j++){
			if(j!=1) printf(" ");
			printf("%d", j);
		}
		for(int t=0;t0){
				k = m/sum[i-1];
				
				if(m%sum[i-1]==0) m-=(k-1)*sum[i-1];
				else m-=k*sum[i-1], k++;
				
				for(q=1,p=0; p<10;){
					if(mark[p]>0) q++;
					if(q>k) break;
					p++;
				}
				printf(" %d", mark[p]);
				mark[p]=-1;
				i--;
			}
		}
		printf("\n");
   }
   return 0;
}