• 题意:两个整数a, b。求出a, a – 1, a – 2……..b +1这些整数能被拆分成多少个素数相乘,把每个的拆分结果相加起来。例如 a = 6, b = 3. 那么结果= 2(4=2*2) + 1(5=5) + 2(6=2*3) = 5
  • 题解:素数筛法,先把每个数能拆分成多少个素数预处理一下,之后用前缀和的思想,相减一下就行了
    • 然而有些人想用阶乘分解公式,但是阶乘分解公式不能够预处理…
  • 错解
    • //#include<bits/stdc++.h>
      #include<algorithm>
      #include <iostream>
      #include  <cstdlib>
      #include  <cstring>
      #include  <cassert>
      #include   <cstdio>
      #include   <vector>
      #include   <string>
      #include    <cmath>
      #include    <queue>
      #include    <stack>
      #include      <set>
      #include      <map>
      using namespace std;
      #define P(a,b,c) make_pair(a,make_pair(b,c))
      #define rep(i,a,n) for (int i=a;i<=n;i++)
      #define per(i,a,n) for (int i=n;i>=a;i--)
      #define CLR(vis) memset(vis,0,sizeof(vis))
      #define MST(vis,pos) memset(vis,pos,sizeof(vis))
      #define pb push_back
      #define mp make_pair
      #define all(x) (x).begin(),(x).end()
      #define fi first
      #define se second
      #define SZ(x) ((int)(x).size())
      typedef pair<int,pair<int,int> >pii;
      typedef long long ll;
      typedef unsigned long long ull;
      const ll mod = 1000000007;
      const int INF = 0x3f3f3f3f;
      ll gcd(ll a, ll b) {
      	return b ? gcd(b, a%b) : a;
      }
      template<class T>inline void gmax(T &A, T B) {
      	(A<B)&&(A=B);//if(B>A)A=B;
      }
      template<class T>inline void gmin(T &A, T B) {
      	(A>B)&&(A=B);//if(B<A)A=B;
      }
      template <class T>
      inline bool scan_d(T &ret) {
      	char c;
      	int sgn;
      	if(c=getchar(),c==EOF) return 0; //EOF
      	while(c!='?'&&(c<'0'||c>'9')) c=getchar();
      	sgn=(c=='?')?-1:1;
      	ret=(c=='?')?0:(c-'0');
      	while(c=getchar(),c>='0'&&c<='9') ret=ret*10+(c-'0');
      	ret*=sgn;
      	return 1;
      }
      inline void outt(int x) {
      	if(x>9) outt(x/10);
      	putchar(x%10+'0');
      }
      const int maxn=5e6+10;
      int vis[maxn];
      int P[maxn];
      int cnt;
      void init(int n) {
      	vis[1]=1;
      	cnt=0;
      	rep(i,1,n) {
      		if(!vis[i]) {
      			P[++cnt]=i;
      			for(int j=i*2; j<=n; j+=i) {
      				if(!vis[j])vis[j]=1;
      			}
      		}
      	}
      }
      
      
      
      int main() {
      	ios::sync_with_stdio( false );
      	int t;
      	cin>>t;
      	init(5000000);
      	while(t--) {
      		int a,b;
      		cin>>a>>b;
      		ll ans=0;
      		rep(i,1,cnt){
      			ll base=P[i];
      			if(base>a)break;
      			ll num=base;
      			while(num<=a){
      				ans+=a/num;
      				if(num<=b)ans-=b/num;
      				num*=base;
      			}
      		}
      		cout<<ans<<endl;
      	}
      	return 0;
      }
  • 正解
    • //#include<bits/stdc++.h>
      #include<algorithm>
      #include <iostream>
      #include  <cstdlib>
      #include  <cstring>
      #include  <cassert>
      #include   <cstdio>
      #include   <vector>
      #include   <string>
      #include    <cmath>
      #include    <queue>
      #include    <stack>
      #include      <set>
      #include      <map>
      using namespace std;
      #define P(a,b,c) make_pair(a,make_pair(b,c))
      #define rep(i,a,n) for (int i=a;i<=n;i++)
      #define per(i,a,n) for (int i=n;i>=a;i--)
      #define CLR(vis) memset(vis,0,sizeof(vis))
      #define MST(vis,pos) memset(vis,pos,sizeof(vis))
      #define pb push_back
      #define mp make_pair
      #define all(x) (x).begin(),(x).end()
      #define fi first
      #define se second
      #define SZ(x) ((int)(x).size())
      typedef pair<int,pair<int,int> >pii;
      typedef long long ll;
      typedef unsigned long long ull;
      const ll mod = 1000000007;
      const int INF = 0x3f3f3f3f;
      ll gcd(ll a, ll b) {
      	return b ? gcd(b, a%b) : a;
      }
      template<class T>inline void gmax(T &A, T B) {
      	(A<B)&&(A=B);//if(B>A)A=B;
      }
      template<class T>inline void gmin(T &A, T B) {
      	(A>B)&&(A=B);//if(B<A)A=B;
      }
      template <class T>
      inline bool scan_d(T &ret) {
      	char c;
      	int sgn;
      	if(c=getchar(),c==EOF) return 0; //EOF
      	while(c!='?'&&(c<'0'||c>'9')) c=getchar();
      	sgn=(c=='?')?-1:1;
      	ret=(c=='?')?0:(c-'0');
      	while(c=getchar(),c>='0'&&c<='9') ret=ret*10+(c-'0');
      	ret*=sgn;
      	return 1;
      }
      inline void outt(int x) {
      	if(x>9) outt(x/10);
      	putchar(x%10+'0');
      }
      const int maxn=5e6+10;
      bool is_prime[maxn];
      int num[maxn];
      
      void fac(){
      	CLR(num);
      	CLR(is_prime);
      	rep(i,2,maxn-10){
      		if(!is_prime[i]){
      			for(int j=i;j<=maxn-10;j+=i){
      				int tmp=j;
      				while(tmp%i==0){
      					num[j]++;
      					tmp/=i;
      				}
      				if(j!=i)is_prime[j]=1;
      			}
      		}
      	}
      	rep(i,2,maxn-10){
      		num[i]=num[i]+num[i-1];
      	}
      }
      //即使不输入到缓存区,cincout还是TLE
      int main(){
      	//ios::sync_with_stdio( false );
      	fac();
      	int t,a,b;
      	//cin>>t;
      	scan_d(t);
      	while(t--){
      		scan_d(a);
      		scan_d(b);
      		//cin>>a>>b;
      		//cout<<num[a]-num[b]<<"\n";
      		outt(num[a]-num[b]);
      		printf("\n");
      	}
      	return 0;
      }

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