此坑待補
4383: n % ( pow( p , 2) ) ===0 
Time Limit(Common/Java):10000MS/30000MS Memory Limit:65536KByteTotal Submit: 238 Accepted:54
Description
There is a number n , determine whether there is a p (p>1) that p^2 is a divisor of n.
Input
The first line contains an integer T , the number of test case.
The following T lines , each contains an integer n.
( 1<= T <=10^2 , 1<= n <=10^18 )
Output
A integer p , if there exist multiple answer ,output the minimum one.
Or print “oh,no.” .
Sample Input
3
8
16
17
Sample Output
2
2
oh,no.
Source
數信學院第5屆新生程序設計競賽
我的代碼
#include<iostream> #include<cstdio> #include<cstdlib> #include<ctime> #include<algorithm> #include<cstring> using namespace std; typedef __int64 ll; const int times=20; const int N=100; ll mult_mod(ll a,ll b,ll mod) { a%=mod; b%=mod; ll res=0; while(b) { if(b&1) { res+=a; res%=mod; } a<<=1; if(a>=mod) a%=mod; b>>=1; } return res; } ll pow_mod(ll x,ll n,ll mod) { if(n==1) return x%mod; x%=mod; ll t=x; ll res=1; while(n) { if(n&1) res=mult_mod(res,t,mod); t=mult_mod(t,t,mod); n>>=1; } return res; } bool test(ll a,ll n,ll x,ll t) { ll res=pow_mod(a,x,n); ll last=res; for(int i=1; i<=t; i++) { res=mult_mod(res,res,n); if(res==1&&last!=1&&last!=n-1) return true; last=res; } if(res!=1) return true; return false; } bool miller_rabin(ll n) { if(n<2) return false; if(n==2) return true; if((n&1)==0) return false; ll x=n-1,t=0; while((x&1)==0) { x>>=1; t++; } for(int i=0; i<times; i++) { ll a=rand()%(n-1)+1; if(test(a,n,x,t)) return false; } return true; } ll factor[N]; int tot; ll gcd(ll a,ll b) { if(a==0) return 1; if(a<0) return gcd(-a,b); while(b) { ll c=a%b; a=b; b=c; } return a; } ll pollard_rho(ll x,ll c) { ll i=1,k=2; ll x0=rand()%x; ll y=x0; while(1) { i++; x0=(mult_mod(x0,x0,x)+c)%x; ll d=gcd(y-x0,x); if(d!=1&&d!=x) return d; if(y==x0) return x; if(i==k) { y=x0; k+=k; } } } void find_factor(ll n) { if(miller_rabin(n)) { factor[tot++]=n; return ; } ll p=n; while(p>=n) p=pollard_rho(p,rand()%(n-1)+1); find_factor(p); find_factor(n/p); } int main() { srand(time(0)); int t; scanf("%d",&t); ll n; while(t--) { scanf("%I64d",&n); if(miller_rabin(n)||n<=1) { printf("oh,no.\n"); continue; } tot=0; find_factor(n); sort(factor,factor+tot); int f=1; for(int i=1; i<tot; i++) if(factor[i]==factor[i-1]) { printf("%I64d\n",factor[i]); f=0; break; } if(f)printf("oh,no.\n"); } return 0; }
2017: N-Credible Mazes
Time Limit(Common/Java):1000MS/10000MS Memory Limit:65536KByteTotal Submit: 6 Accepted:0
Description
An n-tersection is defined as a location in n-dimensional space, n being a positive integer, having all non-negative integer coordinates. For example, the location (1,2,3) represents an n-tersection in three dimensional space. Two n-tersections are said to be adjacent if they have the same number of dimensions and their coordinates differ by exactly 1 in a single dimension only. For example, (1,2,3) is adjacent to (0,2,3) and (2,2,3) and (1,2,4), but not to (2,3,3) or (3,2,3) or (1,2). An n-teresting space is defined as a collection of paths between adjacent n-tersections.
Finally, an n-credible maze is defined as an n-teresting space combined with two specific n-tersections in that space, one of which is identified as the starting n-tersection and the other as the ending n-tersection.
Input
The input file will consist of the descriptions of one or more n-credible mazes. The first line of the description will specify n, the dimension of the n-teresting space. (For this problem, n will not exceed 10, and all coordinate values will be less than 10.) The next line will contain 2n non-negative integers, the first n of which describe the starting n-tersection, least dimension first, and the next n of which describe the ending n-tersection. Next will be a nonnegative number of lines containing 2n non-negative integers each, identifying paths between adjacent n-tersections in the n-teresting space. The list is terminated by a line containing only the value ?C1. Several such maze descriptions may be present in the file. The end of the input is signalled by space dimension of zero. No further data will follow this terminating zero.
Output
For each maze output it‘s position in the input; e.g. the first maze is "Maze #1", the second is "Maze #2", etc. If it is possible to travel through the n-credible maze‘s n-teresting space from the starting n-tersection to the ending n-tersection, also output "can be travelled" on the same line. If such travel is not possible, output "cannot be travelled" instead.
Sample Input
2
0 0 2 2
0 0 0 1
0 1 0 2
0 2 1 2
1 2 2 2
-1
3
1 1 1 1 2 3
1 1 2 1 1 3
1 1 3 1 2 3
1 1 1 1 1 0
1 1 0 1 0 0
1 0 0 0 0 0
-1
0
Sample Output
Maze #1 can be travelled
Maze #2 cannot be travelled
爆搜的代碼
#include<stdio.h> #include<algorithm> using namespace std; int a[1005][10],b[1005][10],c[1005][3]; int main() { int n,T=1; while(~scanf("%d",&n),n) { int i,j,k; for(i=0;;i++) { scanf("%d",&a[i][0]); if(a[i][0]==-1)break; for(j=1; j<n; j++) scanf("%d",&a[i][j]); for(j=0; j<n; j++) scanf("%d",&b[i][j]); } for(k=0; k<i; k++) for(j=0; j<n; j++) c[k][j]=0; c[0][0]=1; for(i=0; i<k; i++) { int f=0; for(j=0; j<n; j++) if(a[i+1][j]!=a[i][j]) { f++; if(abs(a[i+1][j]-a[i][j])!=1)break; } if(f<=1&&j==n&&c[i][0]==1)c[i+1][0]=1; f=0; for(j=0; j<n; j++) if(b[i][j]!=a[i][j]) { f++; if(abs(b[i][j]-a[i][j])!=1)break; } if(f<=1&&j==n&&c[i][0]==1)c[i][1]=1; f=0; for(j=0; j<n; j++) if(b[i+1][j]!=b[i][j]) { f++; if(abs(b[i+1][j]-b[i][j])!=1)break; } if(f<=1&&j==n&&c[i][1]==1)c[i+1][1]=1; f=0; for(j=0; j<n; j++) if(b[i+1][j]!=a[i+1][j]) { f++; if(abs(b[i+1][j]-a[i+1][j])!=1)break; } if(f<=1&&j==n&&c[i+1][1]==1)c[i+1][0]=1; } if(c[k-1][1]>=1) printf("Maze #%d can be travelled\n",T++); else printf("Maze #%d cannot be travelled\n",T++); } return 0; }
map+set維護
#include<stdio.h> #include<vector> #include<set> #include<map> using namespace std; map<int,vector<int> >M; set<int>S; int la(int &num,int n) { num=0; for(int i=1; i<=n; i++) { int x; scanf("%d",&x); if(x==-1)return 0; num=num*10+x; } return 1; } int dfs(int st,int ed) { if(st==ed)return 1; int l=M[st].size(); for(int i=0; i<l; i++) { int v=M[st][i]; if(!S.count(v)) { S.insert(v); if(dfs(v,ed))return 1; } } return 0; } int main() { int t=1,n; while(~scanf("%d",&n)) { if(n==0) { if(scanf("%d",&n)==EOF) break; } M.clear(),S.clear(); int st,ed; la(st,n); la(ed,n); int s,e; while(la(s,n)) { la(e,n); M[s].push_back(e); M[e].push_back(s); } if(dfs(st,ed)) printf("Maze #%d can be travelled\n",t++); else printf("Maze #%d cannot be travelled\n",t++); } return 0; }
網上的HDUac代碼
#include<stdio.h> #include<string.h> int f,sum,e,s; int a[10010],b[10010]; int mark[10010]; void dfs(int w) { int i; if (w==e) { f=1; return ; } for (i=1; i<=sum; i++) if (mark[i]==0 && (a[i]==w || b[i]==w)) { mark[i]=1; if (a[i]==w) dfs(b[i]); else dfs(a[i]); } } int main() { int Case,i,j,n,x; Case=0; while (scanf("%d",&n)!=EOF) { if (n==0) break; Case++; s=e=0; for (i=1; i<=n; i++) { scanf("%d",&x); s=s*10+x; } for (i=1; i<=n; i++) { scanf("%d",&x); e=e*10+x; } sum=0; memset(a,0,sizeof(a)); memset(b,0,sizeof(b)); while (scanf("%d",&x)!=EOF) { if (x==-1) break; sum++; a[sum]=x; for (i=1; i<n; i++) { scanf("%d",&x); a[sum]=a[sum]*10+x; } for (i=1; i<=n; i++) { scanf("%d",&x); b[sum]=b[sum]*10+x; } } f=0; memset(mark,0,sizeof(mark)); dfs(s); printf("Maze #%d ",Case); if (f==0) printf("cannot be travelled\n"); else printf("can be travelled\n"); } return 0; }
此坑待補