传送门什么的@百度。。
Prime Distance Time Limit: 1000MS Memory Limit: 65536K Description The branch of mathematics called number theory is about properties of numbers. One of the areas that has captured the interest of number theoreticians for thousands of years is the question of primality. A prime number is a number that is has no proper factors (it is only evenly divisible by 1 and itself). The first prime numbers are 2,3,5,7 but they quickly become less frequent. One of the interesting questions is how dense they are in various ranges. Adjacent primes are two numbers that are both primes, but there are no other prime numbers between the adjacent primes. For example, 2,3 are the only adjacent primes that are also adjacent numbers. Your program is given 2 numbers: L and U (1<=L< U<=2,147,483,647), and you are to find the two adjacent primes C1 and C2 (L<=C1< C2<=U) that are closest (i.e. C2-C1 is the minimum). If there are other pairs that are the same distance apart, use the first pair. You are also to find the two adjacent primes D1 and D2 (L<=D1< D2<=U) where D1 and D2 are as distant from each other as possible (again choosing the first pair if there is a tie). Input Each line of input will contain two positive integers, L and U, with L < U. The difference between L and U will not exceed 1,000,000. Output For each L and U, the output will either be the statement that there are no adjacent primes (because there are less than two primes between the two given numbers) or a line giving the two pairs of adjacent primes. Sample Input 2 17 14 17 Sample Output 2,3 are closest, 7,11 are most distant. There are no adjacent primes. Source Waterloo local 1998.10.17
The Euler function Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Problem Description The Euler function phi is an important kind of function in number theory, (n) represents the amount of the numbers which are smaller than n and coprime to n, and this function has a lot of beautiful characteristics. Here comes a very easy question: suppose you are given a, b, try to calculate (a)+ (a+1)+....+ (b) Input There are several test cases. Each line has two integers a, b (2<a<b<3000000). Output Output the result of (a)+ (a+1)+....+ (b) Sample Input 3 100 Sample Output 3042 Source 2009 Multi-University Training Contest 1 - Host by TJU
后缀数组暂时放了。。因为感觉复赛回来之后代码也就忘差不多了。。？
写的都是O(n)的筛法。。欧拉函数的写法是贾教的。。
codes:
1 #include<set> 2 #include<map> 3 #include<cmath> 4 #include<queue> 5 #include<cstdio> 6 #include<cstdlib> 7 #include<cstring> 8 #include<iostream> 9 #include<algorithm> 10 using namespace std; 11 const int N = 500000*2; 12 #define rep(i,n) for(int i=0;i<n;i++) 13 #define Rep(i,n) for(int i=1;i<=n;i++) 14 #define For(i,l,r) for(int i=l;i<=r;i++) 15 int prime[N+10],l,r,primes[N+10],last,now,ansmin,ansmax; 16 int ans1,ans2,ans3,ans4; 17 bool check[N+10],checks[N+10]; 18 19 void PRIME(int n){ 20 For(i,2,n){ 21 if(!check[i]) prime[++prime[0]]=i; 22 Rep(j,prime[0]){ 23 if(prime[j]*i>n) break; 24 check[prime[j]*i]=true; 25 if(!(i%prime[j])) break; 26 } 27 } 28 } 29 30 void query(int l,int r){ 31 memset(checks,false,sizeof(checks)); 32 primes[0]=0;ansmin=1e9;ansmax=0; 33 ans1=ans2=ans3=ans4=last=now=0; 34 Rep(i,prime[0]){ 35 For(j,l/prime[i],r/prime[i]){ 36 if(prime[i]*j<l||j==1) continue; 37 checks[prime[i]*j-l]=true; 38 } 39 } 40 if(l==1) checks[0]=true; 41 rep(i,r-l+1) 42 if(!checks[i]) { 43 primes[++primes[0]]=i+l; 44 now=i+l; 45 if(now-last>ansmax && last){ 46 ansmax=now-last; 47 ans1=last;ans2=now; 48 } 49 if(now-last<ansmin){ 50 ansmin=now-last; 51 ans3=last;ans4=now; 52 } 53 last=now; 54 } 55 if(primes[0]>1) printf("%d,%d are closest, %d,%d are most distant. ",ans3,ans4,ans1,ans2); 56 else puts("There are no adjacent primes."); 57 } 58 59 int main(){ 60 PRIME(N); 61 while(scanf("%d%d",&l,&r)!=EOF) query(l,r); 62 return 0; 63 }
1 #include<set> 2 #include<map> 3 #include<queue> 4 #include<cstdio> 5 #include<cstdlib> 6 #include<cstring> 7 #include<iostream> 8 #include<algorithm> 9 using namespace std; 10 const int N = 3000000; 11 #define rep(i,n) for(int i=0;i<n;i++) 12 #define Rep(i,n) for(int i=1;i<=n;i++) 13 #define For(i,l,r) for(int i=l;i<=r;i++) 14 15 int a,b; 16 int phi[N+3],prime[1000000]; 17 bool check[N+3]; 18 19 void PHI(int n){ 20 phi[1]=1; 21 For(i,2,n){ 22 if(!check[i]){ 23 prime[++prime[0]]=i; 24 phi[i]=i-1; 25 } 26 Rep(j,prime[0]){ 27 if(prime[j]*i>n) break; 28 check[prime[j]*i]=true; 29 if(i%prime[j]) phi[i*prime[j]]=phi[i]*(prime[j]-1); 30 else { 31 phi[i*prime[j]]=phi[i]*prime[j]; 32 break; 33 } 34 } 35 } 36 } 37 38 int main(){ 39 PHI(N); 40 while(scanf("%d%d",&a,&b)!=EOF) { 41 long long sum=0; 42 For(i,a,b) sum+=phi[i]; 43 printf("%I64d ",sum); 44 } 45 return 0; 46 }