Hdu 5943 (prime interval + binary graph matching)

Kingdom of Obsession

Time limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)
Total submission (s): Accepted submission (s): 64

Problem Descriptionthere is a kindom of obsession, so people in this kingdom do things very strictly.

They name themselves in integer, and there isNPeople with their ID continuous(S+1,S+2,? ,s +n) Standing in a line in arbitrary order, being more obsessively, people with IDxWants to stand at yth Position which satisfy

X mody< Span id= "mathjax-span-41" class= "Mo" style= "font-family:mathjax_main; Padding-left:0.278em ">= 0

Is there any satisfy everyone ' s requirement?

Inputfirst line contains an integerT, which indicates the number of test cases.

Every test case contains one line with the integersN,s.

Limits
1≤T≤ .
1≤n≤9 .
0≤s≤9 .

Outputfor every test case, should output ' case #x: Y ', where xIndicates the case number and counts from 1and yis the result string.

If there is an any-to-satisfy everyone's requirement, yequals ' Yes ', otherwise yequals ' No '.

Sample INPUT25 144 11

Sample outputcase #1: nocase #2: Yes

Source 2016 China University student Program Design Competition (Hangzhou) When I think about the number of prime number two times can not match, but did not think of prime spacing (the subconscious thought that the prime number within 2 billion is definitely spaced very large). 2 billion The maximum interval between two primes will not exceed 300, So more than 600 direct output No. So the solution is the small data binary map matching, big data directly output No. To prevent the intersection of the interval, we should give a special n,s size.

#include <stdio.h>#include<algorithm>#include<string.h>#include<math.h>using namespaceStd;typedefLong LongLL;Const intN = .;intS,n;intGraph[n][n];intLinker[n];BOOLVis[n];BOOLDfsintu) {     for(intI=1; i<=n;i++){        if(graph[u][i]==1&&!Vis[i]) {Vis[i]=true; if(linker[i]==-1||DFS (Linker[i])) {Linker[i]=u; return true; }        }    }    return false;}intMain () {intt =1, Tcase; scanf ("%d",&tcase);  while(tcase--) {scanf ("%d%d",&s,&N); if(s<N) swap (s,n); if(n> +) {printf ("Case #%d:no\n", t++); Continue; } memset (graph,0,sizeof(graph)); intres =0;  for(intI=1; i<=n;i++){            intt = i+s;  for(intj=1; j<=n;j++){                if(t%j==0) {Graph[i][j]=1; }}} memset (linker,-1,sizeof(linker));  for(intI=1; i<=n;i++) {memset (Vis,0,sizeof(VIS)); if(Dfs (i)) res++; }        if(res==N) {printf ("Case #%d:yes\n", t++); }Elseprintf"Case #%d:no\n", t++); }    return 0;}

Hdu 5943 (prime interval + binary graph match)