#include #include #include #include #include #include #include #include #include #include #include #include<map name="name" id="name"><div> using namespace std; const int INF=0xfff,MAXN=10000; int PRIME[MAXN]={0},QUAD[MAXN]={0}; typedef long long int LL; int TOP[MAXN][MAXN]; inline int gcd(int a,int b){ //Use Euclidean Algorithm to figure out the Greatest Common Divisor(GCD). //用欧几里得算法求出最大公约数 if(!b) return a; else return gcd(b,a%b); } inline int lcm(int a,int b){ //Figure out the Least Common Multiple(LCM). //求出最小公倍数 return a/gcd(a,b)*b; } inline int kgcd(int &x,int &y,int &d,int a,int b){ //Expand Euclidean Algorithm //I cannot use it very well. //拓展欧几里得 //我不太会用 if(!b) { d=a;x=1;y=0;return a; } else { return kgcd(y,x,d,b,a%b); y-=x*a/b; } } inline int POM(int a,int b,int c){ //Fast Power Modulus //快速幂取模 int ans=1; while(b>0){ if(b%2==1) ans=(ans*a)%c; b=b/2; a=(a*a)%c; } return ans; } inline int LIS(int a[],int len,int *t){ //Get the length of the Longest Increase Sequence and return one of them. //求最长上升子序列长度，返回其中一条子序列 int r[MAXN],tot=-1,d[MAXN]; r[0]=1; for(int i=0;im) tot=j; m=max(m,r[j]); } r[i]=m+1; d[i]=tot; } int ans=0; for(int i=0;ians) tot=i; ans=max(ans,r[i]); } for(int i=ans-1;i>=0;i--){ t[i]=a[tot]; tot=d[tot]; } return ans; } inline int LDS(int a[],int len){ //Figure out the length of the Longest Decline Sequence //求出最长下降子序列长度 int r[MAXN]; r[0]=1; for(int i=0;ia[i]) // If it requires not rise, add '=' // 如果是不升 加上 = 号 m=max(m,r[j]); r[i]=m+1; } int ans=0; for(int i=0;iv[i-1][j]){ v[i][j]=v[i][j-1]; b[i][j]=2; } else if(v[i][j-1]=n) QUAD[j]=1; } return; } inline bool Is_prime(int num){ //Return true: Prime ,or not//num have to be smaller than m //返回真为素数 否则为合数//num必须比m小 if(PRIME[num]==0) return true; else return false; } inline bool Is_quadrat(int num,int n,int m){ //Return false: Quadrat-free,or not//num have to be in [n,m] //返回假为无平方因子数，否则不是//num必须在[n,m]区间内 if(QUAD[num-n]==0) return true; else return false; } inline int Num_prime(int n,int m){ //Figure out the number of the primes in [n,m] //求出在[n,m]区间里的素数个数 int ans=0; for(int i=n;i<=m;i++) if(PRIME[i]==0) ans++; return ans; } inline int Num_quadrat(int n,int m){ //Figure out the number of the quadrat-free numbers in [n,m] //求出在[n,m]区间里的无平方因子数个数 int ans=0; for(int i=0;i<=m-n;i++) if(QUAD[i]==0) ans++; return ans; } int main(){ //std::ios::sync_with_stdio(false); freopen("filename.in","r",stdin); freopen("filename.out","w",stdout); //The main fuction examine input and output //主函数检验输入输出 int a[MAXN],l,ans,t[MAXN]; //Longest Rise Sequence scanf("%d",&l); for(int i=0;i<l;i++) scanf("%d",&a[i]); ans=LIS(a,l,t); printf("LIS: %d\n",ans); for(int i=0;i<ans;i++) printf("%d ",t[i]); //Longest Decline Sequence ans=LDS(a,l); printf("\nLDS: %d\n",ans); // int K,M,C; scanf("%d%d%d",&K,&M,&C); ans=POM(K,M,C); printf("POM: %d\n",ans); ans=gcd(K,M); printf("GCD: %d\n",ans); ans=lcm(K,M); printf("LCM: %d\n",ans); char Q[MAXN]={' '},W[MAXN]={' '}; scanf("%s",Q+1); scanf("%s",W+1); ans=LCS(Q,W); printf("LCS: %d\n",ans); int Begin,End; scanf("%d%d",&Begin,&End); Eno(Begin,End); ans=Num_prime(Begin,End); printf("Num_Prime: %d\n",ans); ans=Num_quadrat(Begin,End); printf("Num_Quadrat: %d\n",ans); fclose(stdin); fclose(stdout); return 0; } </div></map>
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