SGU 169
题意：求k位数里面有多少个是完美数，完美数的定义就是n是好数，n+1也是好数，那么n就是完美数，好数就是n%p(n)==0&&p(n)!=0，p(n)=a1*...*an
收获：找规律
#include<bits/stdc++.h> #define de(x) cout<<#x<<"="<<x<<endl; #define dd(x) cout<<#x<<"="<<x<<" "; #define rep(i,a,b) for(int i=a;i<(b);++i) #define repd(i,a,b) for(int i=a;i>=(b);--i) #define repp(i,a,b,t) for(int i=a;i<(b);i+=t) #define ll long long #define mt(a,b) memset(a,b,sizeof(a)) #define fi first #define se second #define inf 0x3f3f3f3f #define INF 0x3f3f3f3f3f3f3f3f #define pii pair<int,int> #define pdd pair<double,double> #define pdi pair<double,int> #define mp(u,v) make_pair(u,v) #define sz(a) (int)a.size() #define ull unsigned long long #define ll long long #define pb push_back #define PI acos(-1.0) #define qc std::ios::sync_with_stdio(false) #define db double #define all(a) a.begin(),a.end() const int mod = 1e9+7; const int maxn = 1e5+5; const double eps = 1e-6; using namespace std; bool eq(const db &a, const db &b) { return fabs(a - b) < eps; } bool ls(const db &a, const db &b) { return a + eps < b; } bool le(const db &a, const db &b) { return eq(a, b) || ls(a, b); } ll gcd(ll a,ll b) { return a==0?b:gcd(b%a,a); }; ll lcm(ll a,ll b) { return a/gcd(a,b)*b; } ll kpow(ll a,ll b) {ll res=1;a%=mod; if(b<0) return 1; for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;} ll read(){ ll x=0,f=1;char ch=getchar(); while (ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while (ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } int bit(ll n){ int ret = 0; while(n) n/=10,ret++; return ret; } //ll p(ll n){ // ll ret = 1; //// dd(n) // while(n) ret*=(n%10),n/=10; //// de(ret) // return ret; //} //bool check(ll n){ // ll tmp = p(n); // if(tmp==0) return false; // if(n%tmp==0) return true; // return false; //} //bool ok(ll n){ // if(check(n)&&check(n+1)) return true; // return false; //} int main(){ // rep(i,1,1e9+1) { // if(ok(i)) { // dd(bit(i))de(i) // } // } qc; int k; cin>>k;--k; if(!k) printf("8"); else { if(k%6==0) printf("4"); else if(k%3==0) printf("3"); else printf("1"); } return 0; }
SGU 172
题意：m个人，每个人选择了两个考试，那么这两个考试不能在同一天考（n个考试），问你能不能满足所有的人
收获：dfs染色
#include<bits/stdc++.h> #define de(x) cout<<#x<<"="<<x<<endl; #define dd(x) cout<<#x<<"="<<x<<" "; #define rep(i,a,b) for(int i=a;i<(b);++i) #define repd(i,a,b) for(int i=a;i>=(b);--i) #define repp(i,a,b,t) for(int i=a;i<(b);i+=t) #define ll long long #define mt(a,b) memset(a,b,sizeof(a)) #define fi first #define se second #define inf 0x3f3f3f3f #define INF 0x3f3f3f3f3f3f3f3f #define pii pair<int,int> #define pdd pair<double,double> #define pdi pair<double,int> #define mp(u,v) make_pair(u,v) #define sz(a) (int)a.size() #define ull unsigned long long #define ll long long #define pb push_back #define PI acos(-1.0) #define qc std::ios::sync_with_stdio(false) #define db double #define all(a) a.begin(),a.end() const int mod = 1e9+7; const int maxn = 2e2+5; const double eps = 1e-6; using namespace std; bool eq(const db &a, const db &b) { return fabs(a - b) < eps; } bool ls(const db &a, const db &b) { return a + eps < b; } bool le(const db &a, const db &b) { return eq(a, b) || ls(a, b); } ll gcd(ll a,ll b) { return a==0?b:gcd(b%a,a); }; ll lcm(ll a,ll b) { return a/gcd(a,b)*b; } ll kpow(ll a,ll b) {ll res=1;a%=mod; if(b<0) return 1; for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;} ll read(){ ll x=0,f=1;char ch=getchar(); while (ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while (ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } //inv[1]=1; //for(int i=2;i<=n;i++) inv[i]=(mod-mod/i)*inv[mod%i]%mod; int n,m,u,v; int a[maxn][maxn]; int col[maxn]; vector<int> ans; void init(){ mt(col,-1);ans.clear(); rep(i,1,201) rep(j,1,201) a[i][j] = 0; } bool dfs(int u,int c){ col[u] = c; rep(i,1,n+1){ if(a[u][i]){ a[u][i]=a[i][u]=0; if(col[i]==-1) { if(!dfs(i,c^1)) return false; } else if(col[i]==c) return false; } } return true; } int main(){ init(); scanf("%d%d",&n,&m); rep(i,0,m) scanf("%d%d",&u,&v),a[u][v]=a[v][u]=1; rep(i,1,n+1) if(col[i]==-1) { if(!dfs(i,0)) return puts("no"),0; } puts("yes"); rep(i,1,n+1) if(col[i]) ans.pb(i); printf("%d ",sz(ans)); rep(i,0,sz(ans)) printf("%d%c",ans[i]," "[i+1==sz(ans)]); return 0; }
今天的SGU炸了，我登不上，决定先去睡觉了，明天至少4题