"Exhaustion" Poj3187-backward Digit Sums

Source: Internet
Author: User

Ideas

Using the Yang Hui triangle, each number is added equal to the number it corresponds to in the Yang Hui Triangle. Note that the meaning of this problem is that the bottom level is the full ordering of N, not the 1. 10 is all you can. When the Yang Hui triangle was generated, the first time I used a two-cycle simulation to generate, later learned that the Yang Hui Triangle, the nth row K number is Cnk. However, in the second program, my Yang Hui triangle is not preprocessed, resulting in a lot of wasted time. Two methods of deep search and STL are used. Deep search is obviously faster than next_permuation because it can be pruned.

1 /*232K 0MS*/ 2 /*simulation + Deep Search*/3#include <iostream>4#include <cstdio>5 using namespacestd;6 Const intmaxn=Ten+5;7 intn,sum;8 intA[MAXN][MAXN];9 intANS[MAXN];Ten intVIS[MAXN]; One intF; A  - voidGettri () - { the      for(intI=0; i<n;i++) -     { -a[i][0]=1; -a[i][i]=1; +     } -      for(intI=2; i<n;i++) +          for(intj=1; j<i;j++) Aa[i][j]=a[i-1][j-1]+a[i-1][j]; at } -  - voidprint () - { -      for(intI=0; i<n;i++) cout<<ans[i]<<' '; -cout<<Endl; in } -  to voidGetnum (intStepintnowsum) + { -     if(step==N) the     { *         if(nowsum==sum) $         {Panax Notoginsengf=1; - printf (); the         }  +         return; A     } the     if(f| | Nowsum>sum)return; +      for(intI=1; i<=n;i++) -     { $         if(vis[i]==1)Continue; $vis[i]=1; -ans[step]=i; -Getnum (step+1, nowsum+i*a[n-1][step]); thevis[i]=0; -     }Wuyi } the  - intMain () Wu { -scanf"%d%d",&n,&sum); About Gettri (); $      -memset (Vis,0,sizeof(Vis)); -f=0; -Getnum (0,0); A     return 0; +}

1 /*232K 469MS*/2 /*Combination +stl*/ 3#include <iostream>4#include <cstdio>5#include <cstring>6#include <algorithm>7 using namespacestd;8 Const intmaxn=Ten+3;9 intn,sum;Ten intANS[MAXN]; One  A intCintNintk) - { -     intcresult=1; the      for(intI=0; i<k;i++) cresult=cresult* (n-i)/(i+1); -     //This cannot be written as cresult=cresult* (n-i)/(k-i), because accuracy can result in errors if the size from large to small may not be divisible -     returnCresult; - } +  - intMain () + { Ascanf"%d%d",&n,&sum); at      for(intI=0; i<n;i++) ans[i]=i+1; -      Do -     { -         intresult=0; -          for(intI=0; i<n;i++) Result+=ans[i]*c (n1, i); -         if(result==sum) in         { -              for(intI=0; i<n;i++) cout<<ans[i]<<' '; tocout<<Endl; +              Break; -         } the} while(Next_permutation (ans,ans+n)); *     return 0; $}

"Exhaustion" Poj3187-backward Digit Sums

Contact Us

The content source of this page is from Internet, which doesn't represent Alibaba Cloud's opinion; products and services mentioned on that page don't have any relationship with Alibaba Cloud. If the content of the page makes you feel confusing, please write us an email, we will handle the problem within 5 days after receiving your email.

If you find any instances of plagiarism from the community, please send an email to: info-contact@alibabacloud.com and provide relevant evidence. A staff member will contact you within 5 working days.

A Free Trial That Lets You Build Big!

Start building with 50+ products and up to 12 months usage for Elastic Compute Service

  • Sales Support

    1 on 1 presale consultation

  • After-Sales Support

    24/7 Technical Support 6 Free Tickets per Quarter Faster Response

  • Alibaba Cloud offers highly flexible support services tailored to meet your exact needs.