int与long long效率问题

实验题

ps：之前听过说long long的时间问题但是一直没有注意到过。直到遇见了他。。。。
这是一个瞎搞题点这里

思路：瞎搞
证明：int快
本着使劲优化的原则，，，加了4个优化。。但是在int数据类型其实这需要加2个多一点的优化就能过去。。。long long下需要加4个完整的优化才可以。
优化1：质数ans=(1+n)/2
优化2：因子只有2个质数组合
优化3：因子只有3个质数组合
优化4：因子只有4个质数组合
每个优化都有对应的ans遍不需要暴力来处理。。。。
但是在4种优化共同起作用时，

int 300ms

long long 900ms

wuwuwu
本来早就可以水过
先附上瞎搞代码
int

#pragma GCC optimize(3,"Ofast","inline")  	//G++
#include
#define mem(a,x) memset(a,x,sizeof(a))
#define debug(x) cout << #x << ": " << x << endl;
#define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fcout cout<
using namespace std;
typedef long long ll;
//======================================
namespace FastIO{
char print_f[105];void read() {}void print() {putchar('\n');}
template <typename T, typename... T2>
inline void read(T &x, T2 &... oth){x = 0;char ch = getchar();int f = 1;while (!isdigit(ch)){if (ch == '-')f *= -1;ch = getchar();}while (isdigit(ch)){x = x * 10 + ch - 48;ch = getchar();}x *= f;read(oth...);}
template <typename T, typename... T2>
inline void print(T x, T2... oth){int p3=-1;if(x<0) putchar('-'),x=-x;do{print_f[++p3] = x%10 + 48;}while(x/=10);while(p3>=0) putchar(print_f[p3--]);putchar(' ');print(oth...);}} // namespace FastIO
using FastIO::print;
using FastIO::read;
//======================================
typedef pair<int,int> pii;
const int inf=0x3f3f3f3f;
const int mod=1e9+7;
const int maxn = 1e6+5;
bool vis[maxn];
int vis1[maxn],vis2[maxn],vis3[maxn];
int prime[maxn],pos;
void oulasai(int n)  //欧拉筛
{
    for(int i=2;i<=n;i++)
    {
        if(!vis[i]) prime[++pos]=i;
        for(int j=1;j<=pos;j++)
        {
            if(i*prime[j]>n) break;
            vis[i*prime[j]]=true;
            if(i%prime[j]==0) break;
        }
    }
    vis[1]=1;
}
int cntt=0;
int solve(int n){
    int st=sqrt(n);
    for(int i=st;i>=1;i--){
        cntt++;
        if(n%i==0){
            return (i+n/i)/2;
        }
    }
    return 0;
}
int get(int a,int b,int c){
    int x=0,y=0;
    if(a+b*c<b+a*c) x=a,y=b*c;
    else x=b,y=a*c;
    if(x+y>c+a*b) x=c,y=a*b;
    return (x+y)/2;
}
int get2(int a,int b,int c,int d){
    int arr[4]={a,b,c,d};
    vector<int>v;
    for(int i=0;i<1<<4;i++){
        int res=1;
        for(int j=0;j<4;j++){
            if(i>>j&1) res*=arr[j];
        }
        v.push_back(res);
    }
    sort(v.begin(),v.end());
    return (v[7]+v[8])/2;
}
int res[maxn];
main() {
#ifndef ONLINE_JUDGE
    freopen("H:\\code\\in.in", "r", stdin);
    freopen("H:\\code\\out.out", "w", stdout);
    clock_t c1 = clock();
#endif
//**************************************
    oulasai(1000002);
    for(int i=1;i<=pos;i++){
        for(int j=1;j<=pos&&1ll*prime[i]*prime[j]<=1000000;j++){
            if(1ll*prime[i]*prime[j]<=1000000) vis1[prime[i]*prime[j]]=(prime[i]+prime[j])/2;
        }
    }
    for(int i=1;i<=pos;i++){
        for(int j=i;j<=pos&&1ll*prime[i]*prime[j]<=1000000;j++){
            for(int k=j;k<=pos&&1ll*prime[i]*prime[j]*prime[k]<=1000000;k++)
            if(1ll*prime[i]*prime[j]*prime[k]<=1000000) vis2[prime[i]*prime[j]*prime[k]]=get(prime[i],prime[j],prime[k]);
        }
    }
        for(int i=1;i<=pos;i++){
        for(int j=i;j<=pos&&1ll*prime[i]*prime[j]<=1000000;j++){
            for(int k=j;k<=pos&&1ll*prime[i]*prime[j]*prime[k]<=1000000;k++)
            for(int l=k;l<=pos&&1ll*prime[i]*prime[j]*prime[k]*prime[l]<=1000000;l++)
                if(1ll*prime[i]*prime[j]*prime[k]*prime[l]<=1000000) vis3[prime[i]*prime[j]*prime[k]*prime[l]]=get2(prime[i],prime[j],prime[k],prime[l]);
        }
    }
    int ans=0;
    for(int i=1;i<=1000000;i++){
        if(!vis[i]) ans=ans+(1+i)/2;
        else if(vis1[i]) ans=ans+vis1[i];
        else if(vis2[i]) ans=ans+vis2[i];
        else if(vis3[i]) ans=ans+vis3[i];
        else ans=ans+solve(i);
        ans%=mod;
        res[i]=ans;
    }
    int T;
    read(T);
    while(T--){
        int n;
        read(n);
        print(res[n]);
    }
//**************************************
    
#ifndef ONLINE_JUDGE
    cerr << "Time:" << clock() - c1 << "ms" << endl;
#endif
    return 0;
}

long long

#pragma GCC optimize(3,"Ofast","inline")    //G++
#include
#define mem(a,x) memset(a,x,sizeof(a))
#define debug(x) cout << #x << ": " << x << endl;
#define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fcout cout<
using namespace std;
typedef long long ll;
//======================================
namespace FastIO{
char print_f[105];void read() {}void print() {putchar('\n');}
template <typename T, typename... T2>
inline void read(T &x, T2 &... oth){x = 0;char ch = getchar();ll f = 1;while (!isdigit(ch)){if (ch == '-')f *= -1;ch = getchar();}while (isdigit(ch)){x = x * 10 + ch - 48;ch = getchar();}x *= f;read(oth...);}
template <typename T, typename... T2>
inline void print(T x, T2... oth){ll p3=-1;if(x<0) putchar('-'),x=-x;do{print_f[++p3] = x%10 + 48;}while(x/=10);while(p3>=0) putchar(print_f[p3--]);putchar(' ');print(oth...);}} // namespace FastIO
using FastIO::print;
using FastIO::read;
//======================================
typedef pair<ll,ll> pii;
const ll inf=0x3f3f3f3f;
const ll mod=1e9+7;
const ll maxn = 1e6+5;
bool vis[maxn];
ll vis1[maxn],vis2[maxn],vis3[maxn];
ll prime[maxn],pos;
void oulasai(ll n)  //欧拉筛
{
    for(ll i=2;i<=n;i++)
    {
        if(!vis[i]) prime[++pos]=i;
        for(ll j=1;j<=pos;j++)
        {
            if(i*prime[j]>n) break;
            vis[i*prime[j]]=true;
            if(i%prime[j]==0) break;
        }
    }
    vis[1]=1;
}
int cntt=0;
ll solve(ll n){
    ll st=sqrt(n);
    for(ll i=st;i>=1;i--){
        cntt++;
        if(n%i==0){
            return (i+n/i)/2;
        }
    }
    return 0;
}
ll get(ll a,ll b,ll c){
    ll x=0,y=0;
    if(a+b*c<b+a*c) x=a,y=b*c;
    else x=b,y=a*c;
    if(x+y>c+a*b) x=c,y=a*b;
    return (x+y)/2;
}
ll get2(ll a,ll b,ll c,ll d){
    ll arr[4]={a,b,c,d};
    vector<ll>v;
    for(ll i=0;i<1<<4;i++){
        ll res=1;
        for(ll j=0;j<4;j++){
            if(i>>j&1) res*=arr[j];
        }
        v.push_back(res);
    }
    sort(v.begin(),v.end());
    return (v[7]+v[8])/2;
}
ll res[maxn];
main() {
#ifndef ONLINE_JUDGE
    freopen("H:\\code\\in.in", "r", stdin);
    freopen("H:\\code\\out.out", "w", stdout);
    clock_t c1 = clock();
#endif
//**************************************
    oulasai(1000002);
    for(ll i=1;i<=pos;i++){
        for(ll j=1;j<=pos&&prime[i]*prime[j]<=1000000;j++){
            if(prime[i]*prime[j]<=1000000) vis1[prime[i]*prime[j]]=(prime[i]+prime[j])/2;
        }
    }
    for(ll i=1;i<=pos;i++){
        for(ll j=i;j<=pos&&prime[i]*prime[j]<=1000000;j++){
            for(ll k=j;k<=pos&&prime[i]*prime[j]*prime[k]<=1000000;k++)
            if(prime[i]*prime[j]*prime[k]<=1000000) vis2[prime[i]*prime[j]*prime[k]]=get(prime[i],prime[j],prime[k]);
        }
    }
        for(ll i=1;i<=pos;i++){
        for(ll j=i;j<=pos&&prime[i]*prime[j]<=1000000;j++){
            for(ll k=j;k<=pos&&prime[i]*prime[j]*prime[k]<=1000000;k++)
            for(ll l=k;l<=pos&&prime[i]*prime[j]*prime[k]*prime[l]<=1000000;l++)
                if(prime[i]*prime[j]*prime[k]*prime[l]<=1000000) vis3[prime[i]*prime[j]*prime[k]*prime[l]]=get2(prime[i],prime[j],prime[k],prime[l]);
        }
    }
    ll ans=0;
    for(ll i=1;i<=1000000;i++){
        if(!vis[i]) ans=ans+(1+i)/2;
        else if(vis1[i]) ans=ans+vis1[i];
        else if(vis2[i]) ans=ans+vis2[i];
        else if(vis3[i]) ans=ans+vis3[i];
        else ans=ans+solve(i);
        ans%=mod;
        res[i]=ans;
    }
    ll T;
    read(T);
    while(T--){
        ll n;
        read(n);
        print(res[n]);
    }
//**************************************
     
#ifndef ONLINE_JUDGE
    cerr << "Time:" << clock() - c1 << "ms" << endl;
#endif
    return 0;
}

结论

涨姿势了