转自 stven_king的博客 这是一个求一个排序的下一个排列的函数,可以遍历全排列,要包含头文件<algorithm>下面是以前的笔记 (1) int 类型的next_permutation int main(){ int a[3];a[0]=1;a[1]=2;a[2]=3; do {cout<<a[0]<<" "<<a[1]<<" "<<a[2]<<endl;} while…
这是一个求一个排序的下一个排列的函数,可以遍历全排列,要包含头文件<algorithm>下面是以前的笔记 与之完全相反的函数还有prev_permutation (1) int 类型的next_permutation int main(){ int a[3];a[0]=1;a[1]=2;a[2]=3; do{cout<<a[0]<<" "<<a[1]<<" "<<a[2]<<…
在头文件<algorithm>里面有如下代码: int a[]; do { } while(next_permutation(a,a+n)); 可产生1~n的全排列有如下代码: #include <stdio.h> #include <algorithm> using namespace std; int main(){ int n; while(scanf("%d",&n)&&n){ ]; ;i<n;i++){ sca…
Ignatius and the Princess II Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 6191 Accepted Submission(s): 3664 Problem Description Now our hero finds the door to the BEelzebub feng5166. He o…
A - Intersect Until You're Sick of It Time Limit:500MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u Submit Status Practice URAL 2036 Description Ural contests usually contain a lot of geometry problems. Many participants do not conceal…