1001 Gaussian Prime
http://www.tzcoder.cn/acmhome/problemdetail.do?&method=showdetail&id=3798
.......................................我是真的一言难尽
1002 Sum of Factorials
http://www.tzcoder.cn/acmhome/problemdetail.do?&method=showdetail&id=2696
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 #define N 1000005 5 int a[15]; 6 void init() 7 { 8 a[0]=a[1]=1; 9 for(int i=2;i<10;i++) 10 a[i]=a[i-1]*i; 11 } 12 int main() 13 { 14 init(); 15 int n; 16 while(~scanf("%d",&n),n>=0) 17 { 18 if(n==0) 19 { 20 printf("NO "); 21 continue; 22 } 23 int all=0; 24 for(int i=9;i>=0;i--) 25 { 26 all+=a[i]; 27 if(all>n) all-=a[i]; 28 } 29 if(all==n) printf("YES "); 30 else printf("NO "); 31 } 32 }
1003 Billboard
http://www.tzcoder.cn/acmhome/problemdetail.do?&method=showdetail&id=6016
黑板报h*w，第i个广告是1*wi。优先上、左。线段树存剩余最大容量。
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 #define M 200005 5 int tre[M<<2]; 6 void build(int ind,int left,int right,int w) 7 { 8 tre[ind]=w; 9 if(left==right) return; 10 int mid=(left+right)>>1; 11 build(ind*2,left,mid,w); 12 build(ind*2+1,mid+1,right,w); 13 tre[ind]=max(tre[ind*2],tre[ind*2+1]); 14 } 15 int finda(int ind,int left,int right,int a) 16 { 17 if(left==right) 18 { 19 tre[ind]-=a; 20 return left; 21 } 22 int mid=(left+right)>>1,ans; 23 if(tre[ind*2]>=a) ans=finda(ind*2,left,mid,a); 24 else ans=finda(ind*2+1,mid+1,right,a); 25 tre[ind]=max(tre[ind*2],tre[ind*2+1]); 26 return ans; 27 } 28 int main() 29 { 30 int h,w,n; 31 while(~scanf("%d%d%d",&h,&w,&n)) 32 { 33 int minn=min(h,n); 34 build(1,1,minn,w); 35 while(n--) 36 { 37 int a;scanf("%d",&a); 38 if(tre[1]<a) printf("-1 "); 39 else printf("%d ",finda(1,1,minn,a)); 40 } 41 } 42 }
1004 Monkey Party
http://www.tzcoder.cn/acmhome/problemdetail.do?&method=showdetail&id=6054
咕
1005 Happy Necklace
http://www.tzcoder.cn/acmhome/problemdetail.do?&method=showdetail&id=6064
一点都不happy的项链，f(2)=3, f(3)=4, f(4)=6, f(n)=f(n-1)+f(n-2)，矩阵快速幂。
1 1 0
A = 0 0 1
1 0 0
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 const int mod=1e9+7; 5 struct p{ 6 ll rol,col; 7 ll matris[3][3]; 8 }; 9 p Matris(p a,p b) 10 { 11 p tem; 12 memset(tem.matris,0,sizeof(tem.matris)); 13 for(ll i=0;i<3;i++) 14 for(ll j=0;j<3;j++) 15 for(ll k=0;k<3;k++) 16 { 17 tem.matris[i][j]+=a.matris[i][k]*(b.matris[k][j]%mod); 18 tem.matris[i][j]%=mod; 19 } 20 return tem; 21 } 22 ll quick_mi(ll n) 23 { 24 p res,tep; 25 res.matris[0][0]=6,res.matris[0][1]=4,res.matris[0][2]=3; 26 for(int i=1;i<=2;i++) 27 for(int j=0;j<=2;j++) 28 res.matris[i][j]=0; 29 tep.matris[0][0]=1,tep.matris[0][1]=1,tep.matris[0][2]=0; 30 tep.matris[1][0]=0,tep.matris[1][1]=0,tep.matris[1][2]=1; 31 tep.matris[2][0]=1,tep.matris[2][1]=0,tep.matris[2][2]=0; 32 while(n) 33 { 34 if(n&1) res=Matris(res,tep); 35 tep=Matris(tep,tep); 36 n/=2; 37 } 38 return res.matris[0][0]%mod; 39 } 40 int main() 41 { 42 int t;scanf("%d",&t); 43 while(t--) 44 { 45 ll n;scanf("%lld",&n); 46 if(n==2) printf("3 "); 47 else if(n==3) printf("4 "); 48 else if(n==4) printf("6 "); 49 else printf("%lld ",quick_mi(n-4)); 50 } 51 }
1006 CA Loves GCD
http://www.tzcoder.cn/acmhome/problemdetail.do?&method=showdetail&id=6068
我不爱
1007 Squarefree number
http://www.tzcoder.cn/acmhome/problemdetail.do?&method=showdetail&id=6071
欧拉筛把10^6内的素数标下，如果除了两个或以上就No。素数除完了可能还很大，就要看是不是完全平方数，否则n就是一个很大的素数或者两个大于10^6的素数乘积。
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 #define N 1000005 5 int prim[N+5]; 6 int vis[N+5],cnt=0; 7 void init() 8 { 9 vis[0]=vis[1]=1; 10 for(int i=0;i<N;i++) 11 if(!vis[i]) 12 { 13 prim[cnt++]=i; 14 for(ll j=(ll)i*i;j<N;j+=i) vis[j]=1; 15 } 16 } 17 int main() 18 { 19 init(); 20 int t;scanf("%d",&t); 21 for(int kk=1;kk<=t;kk++) 22 { 23 int flag=1; 24 ll n;scanf("%lld",&n); 25 for(int i=0;i<cnt;i++) 26 { 27 if(n%prim[i]==0) 28 { 29 int num=0; 30 while(n%prim[i]==0) n/=prim[i],num++; 31 if(num>=2) 32 { 33 flag=0; 34 printf("Case %d: No ",kk); 35 break; 36 } 37 } 38 } 39 if(n>1000000&&flag) 40 { 41 int q=(int)sqrt(n); 42 if((ll)q*q==n) printf("Case %d: No ",kk); 43 else printf("Case %d: Yes ",kk); 44 continue; 45 } 46 if(flag) printf("Case %d: Yes ",kk); 47 } 48 }
1008 Gym Class
http://www.tzcoder.cn/acmhome/problemdetail.do?&method=showdetail&id=6081
选自己前面包括自己的最小id作为评分，记录一下要加上的评分minn。没有要求的入度为0进队列。
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 const int MAXN=1e9; 5 const int N=100005; 6 vector<int> vec[N]; 7 int inde[N]; 8 priority_queue<int> q; 9 void init() 10 { 11 memset(inde,0,sizeof(inde)); 12 for(int i=1;i<=N;i++) vec[i].clear(); 13 while(!q.empty()) q.pop(); 14 } 15 int main() 16 { 17 int t;scanf("%d",&t); 18 while(t--) 19 { 20 ll ans=0,minn=MAXN; 21 init(); 22 int n,m;scanf("%d%d",&n,&m); 23 while(m--) 24 { 25 int a,b;scanf("%d%d",&a,&b); 26 inde[b]++; 27 vec[a].push_back(b); 28 } 29 for(int i=1;i<=n;i++) 30 if(!inde[i]) q.push(i); 31 while(!q.empty()) 32 { 33 ll p=q.top();q.pop(); 34 minn=min(p,minn); 35 ans+=minn; 36 for(int i=0;i<vec[p].size();i++) 37 { 38 inde[vec[p][i]]--; 39 if(inde[vec[p][i]]==0) q.push(vec[p][i]); 40 } 41 } 42 printf("%lld ",ans); 43 } 44 }