HDU1027 强大的STL

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再一次见识到了STL的强大，，ORZ。。。

eg：b[3]={1,2,3};(有序)

有3!个全排列

next_permutation(b+0, b + 3); (递增)

The 3! possible permutations with 3 elements:1 2 31 3 22 1 32 3 13 1 23 2 1After loop: 1 2 3

相对的有prev_permutation为递减

#include<cstdio>#include<stdlib.h>#include<string.h>#include<string>#include<map>#include<cmath>#include<iostream>#include <queue>#include <stack>#include<algorithm>#include<set>using namespace std;#define inf 2147483647#define eps 1e-8#define LL long long#define M 50005#define mol 1000000007int main(){int n,m;int a[]={0,1,2,6,24,120,720,5040,40320},b[1005]={0};while(~scanf("%d%d",&n,&m)){int i,j,k;   for(i=0;i<10;i++)     if(m>=a[i]&&m<=a[i+1]){ k=i+1;break;   }   for(j=1;j<=n-k;j++)   printf("%d ",j);   for(i=n-k+1,j=0;i<=n;i++,j++)   {      b[j]=i;   }   m--;   while(m--)   {      next_permutation(b+0, b + k);     }   for(i=0;i<j-1;i++)   printf("%d ",b[i]);   printf("%d\n",b[j-1]);}return 0;}