4 Values whose Sum is 0

#include <iostream>
#include <algorithm>
#include <vector>
#include <cstdio>
const int mxn = 5e6+7 ;
using namespace std;
typedef long long ll;

ll p,t,n,m, si , a[mxn] , b[mxn] , c[mxn] , d[mxn] , rc[mxn] ,sum[mxn];
bool ans[mxn];
template <class T>
void rd(T &x){
    x=0;T f=1;char c=getchar();
    while(!isdigit(c)) {if(c=='-') f = -1 ; c = getchar();}
    while(isdigit(c)) {x = (x<<1) + (x<<3) + (c^48);c = getchar();}
    x*=f;
}
int main()
{
    rd(n);
    for(int i=1;i<=n;i++)
        rd(a[i]) , rd(b[i]) , rd(c[i]) , rd(d[i]);
    vector<ll> v1 , v2;
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
            v1.push_back(a[i]+b[j]) , v2.push_back(c[i]+d[j]);
    sort(v1.begin(),v1.end());
    sort(v2.begin(),v2.end());
    int ans = 0 ;
    for(int i=0;i<v1.size();i++){
        int p = lower_bound(v2.begin(),v2.end(),-v1[i]) - v2.begin() ;
        if(v2[p]+v1[i]==0){ /// 计算所有组合
            while(v2[p]+v1[i]==0) ans++ , p++ ;
        }
    }
    printf("%d\n",ans);
}

Pie

#include <iostream>
#include <algorithm>
#include <vector>
#include <cstdio>
#include <cmath>
#define pi acos(-1.0)
#define eps 1e-5
const int mxn = 5e6+7 ;
using namespace std;
typedef long long ll;
template <class T>
void rd(T &x){
    x=0;T f=1;char c=getchar();
    while(!isdigit(c)) {if(c=='-') f = -1 ; c = getchar();}
    while(isdigit(c)) {x = (x<<1) + (x<<3) + (c^48);c = getchar();}
    x*=f;
}
ll p,t,n,m,k,si ;
double a[mxn] ;
bool check(double x)
{
    int ans = 0 ;
    for(int i=1;i<=n;i++){
        ans+= (int)(a[i]/x);
    }
    if(ans>=m+1) return true;
    return false ;
}
int main()
{
    rd(t);
    while(t--){
        rd(n) , rd(m) ;
        for(int i=1;i<=n;i++) 
            rd(k) , a[i] = k*k*pi;
        sort(a+1,a+1+n);
        double l = 0 , r = a[n] ;
        while(r-l>eps){
            double mid = (l+r)/2.0;
            if(check(mid)) l = mid ;
            else r = mid ;
        }
        printf("%.4f\n",r);
    }
}

Can you find it?

#include <iostream>
#include <algorithm>
#include <vector>
#include <cstdio>
#include <cmath>
#define pi acos(-1.0)
#define eps 1e-5
const int mxn = 5e6+7 ;
using namespace std;
typedef long long ll;
template <class T>
void rd(T &x){
    x=0;T f=1;char c=getchar();
    while(!isdigit(c)) {if(c=='-') f = -1 ; c = getchar();}
    while(isdigit(c)) {x = (x<<1) + (x<<3) + (c^48);c = getchar();}
    x*=f;
}
int p,t,n,m,k,si,ans;
int a[mxn] , b[mxn] ,c[mxn];
int main()
{
    int cas = 1 ;
    while(~scanf("%d %d %d",&n,&m,&k)){
        printf("Case %d:\n",cas++ );
        si = 1 ;
        for(int i=1;i<=n;i++) rd(a[i]);
        for(int i=1;i<=m;i++) {
            rd(p);
            for(int j=1;j<=n;j++)
                c[si++] = p+a[j];
        }
        for(int i=1;i<=k;i++) rd(b[i]);
        sort(c+1,c+si);
        rd(ans);
        while(ans--){
            int x , falg = 1 ; rd(x);
            for(int i=1;i<=k;i++){
                int y = lower_bound(c+1,c+si,x-b[i]) - c;

                if( (x == b[i] + c[y] ) && y>=1 ){
                    puts("YES"); falg = 0 ;
                    break;
                }
            }
            if(falg) puts("NO");
        }

    }
}

A Cubic number and A Cubic Number

题解:p 为质数，那么P的因子只有1和p，而 p = a^3 - b^3 = (a-b)(a^2+ab+b^2) = 1(a^2+ab+b^2) = 3b^2 + 3*b + 1

#include <iostream>
#include <algorithm>
#include <vector>
#include <cstdio>
#include <cmath>
#define pi acos(-1.0)
#define eps 1e-5
const int mxn = 5e6+7 ;
using namespace std;
typedef long long ll;
template <class T>
void rd(T &x){
    x=0;T f=1;char c=getchar();
    while(!isdigit(c)) {if(c=='-') f = -1 ; c = getchar();}
    while(isdigit(c)) {x = (x<<1) + (x<<3) + (c^48);c = getchar();}
    x*=f;
}
ll p,t,n,m,k,si,ans;
int a[mxn] , b[mxn] ,c[mxn];
int main()
{
    rd(t);
    while(t--){
        rd(n); ll l = 1 , r = 1000000+7, falg = 1 ;
        while(l<=r){
            ll mid = (l+r)>>1;
            if(1ll*3*mid*mid + 1ll*3*mid + 1==n){
                puts("YES"); falg = 0 ; break;
            }
            if(1ll*3*mid*mid+1ll*3*mid+1>n) r = mid-1 ;
            else l = mid+1 ;
        }
        if(falg) puts("NO");
    }   
}

I Count Two Three

#include <iostream>
#include <algorithm>
#include <vector>
#include <cstdio>
#include <cmath>
#define pi acos(-1.0)
#define eps 1e-5
const int mxn = 1e9+7 ;
const int lim = 1e6+7 ;
using namespace std;
typedef long long ll;
template <class T>
void rd(T &x){
    x=0;T f=1;char c=getchar();
    while(!isdigit(c)) {if(c=='-') f = -1 ; c = getchar();}
    while(isdigit(c)) {x = (x<<1) + (x<<3) + (c^48);c = getchar();}
    x*=f;
}
ll p,t,n,m,k,si,ans;
ll a[lim] ;
int main()
{
    for(ll i=1;i<mxn;i<<=1){
        for(ll j=1;j*i<mxn;j*=3){
            for(ll k=1;k*j*i<mxn;k*=5){
                for(ll x=1;i*j*k*x<mxn;x*=7){
                    a[m++] = i*j*k*x;
                }
            }
        }
    }
    sort(a,a+m);
    rd(n);
    while(n--){
        rd(k);
        printf("%lld\n",*lower_bound(a,a+m,k));
    }
}

Funky Numbers

#include <iostream>
#include <algorithm>
#include <vector>
#include <cstdio>
#include <cmath>
#define pi acos(-1.0)
#define eps 1e-5
const int mxn = 1e9+7 ;
const int lim = 1e6+7 ;
using namespace std;
typedef long long ll;
template <class T>
void rd(T &x){
    x=0;T f=1;char c=getchar();
    while(!isdigit(c)) {if(c=='-') f = -1 ; c = getchar();}
    while(isdigit(c)) {x = (x<<1) + (x<<3) + (c^48);c = getchar();}
    x*=f;
}
ll p,t,n,m,k,si,ans;
ll a[lim] ;
int main()
{
    rd(n); n<<=1;
    int falg = 1;
    for(int i=1;i<=sqrt(n);i++){
        ll ans = n-i*(i+1) ;
        ll l = 1 , r = sqrt(ans) + 1 ;
        while(l<r){
            ll mid = (l+r)>>1;
            if(mid*(mid+1)==ans){
                puts("YES");
                falg = 0 ;
                break;
            } else if(mid*(mid+1)>ans) {
                r = mid - 1 ;
            } else {
                l = mid + 1;
            }
        }
        if(!falg) break;
    }
    if(falg) puts("NO");
}

Burning Midnight Oil

#include <iostream>
#include <algorithm>
#include <vector>
#include <cstdio>
#include <cmath>
#define pi acos(-1.0)
#define eps 1e-5
const int mxn = 1e9+7 ;
const int lim = 1e6+7 ;
using namespace std;
typedef long long ll;
template <class T>
void rd(T &x){
    x=0;T f=1;char c=getchar();
    while(!isdigit(c)) {if(c=='-') f = -1 ; c = getchar();}
    while(isdigit(c)) {x = (x<<1) + (x<<3) + (c^48);c = getchar();}
    x*=f;
}
ll p,t,n,m,k,si,ans;
ll a[lim] ;
bool check(ll x)
{
    ll cnt = x , p = k;
    while(x/p){
        cnt += x/p;
        p*=k;
    }
    if(cnt>=n) return false ;
    return true ;
}
int main()
{
    rd(n) , rd(k);
    ll l = 1 , r = n ;
    while(l<=r){
        ll mid = (r+l)>>1;
        if(check(mid)) l = mid + 1 ;
        else r = mid - 1 ;
    }
    printf("%d\n",l);
}

Can you solve this equation?

#include <iostream>
#include <algorithm>
#include <vector>
#include <cstdio>
#include <cmath>
#define pi acos(-1.0)
#define eps 1e-9
const int mxn = 1e9+7 ;
const int lim = 1e6+7 ;
using namespace std;
typedef long long ll;
template <class T>
void rd(T &x){
    x=0;T f=1;char c=getchar();
    while(!isdigit(c)) {if(c=='-') f = -1 ; c = getchar();}
    while(isdigit(c)) {x = (x<<1) + (x<<3) + (c^48);c = getchar();}
    x*=f;
}
ll p,t,n,m,k,si,ans;
ll a[lim] ;
double check(double x)
{
    return 8*x*x*x*x + 7*x*x*x + 2*x*x + 3*x + 6 ;
}
int main()
{
    rd(t);
    while(t--){
        double n , l = -100 , r = 101 , mid ; 
        scanf("%lf",&n);
        while(r-l>eps){
            mid = (l+r)/2.0;
            if(check(mid)>=n)
                r = mid ;
            else 
                l = mid ;
        }
        if(mid<0 || mid >100){
            puts("No solution!");
        } else {
            printf("%.4f\n",mid);
        }
    }
}

Strange fuction

#include <iostream>
#include <algorithm>
#include <vector>
#include <cstdio>
#include <cmath>
#define pi acos(-1.0)
#define eps 1e-9
const int mxn = 1e9+7 ;
const int lim = 1e6+7 ;
using namespace std;
typedef long long ll;
template <class T>
void rd(T &x){
    x=0;T f=1;char c=getchar();
    while(!isdigit(c)) {if(c=='-') f = -1 ; c = getchar();}
    while(isdigit(c)) {x = (x<<1) + (x<<3) + (c^48);c = getchar();}
    x*=f;
}
ll p,t,n,m,k,si,ans;
ll a[lim] ;
double check(double x) {
    return 42*pow(x,6.0)+48*pow(x,5.0)+21*pow(x,2.0)+10*x;
}
double res(double x,double y){
    return 6*pow(x,7)+8*pow(x,6)+7*pow(x,3)+5*pow(x,2)-x*y;
}
int main()
{
    rd(t); double n;
    while(t--){
        scanf("%lf",&n);
        double l = 0 , r = 101 , mid ;
        while(r-l>eps){
            mid = (l+r)/2.0;
            if(check(mid)>=n)
                r = mid ;
            else 
                l = mid ;
        }
        printf("%.4f\n",res(mid,n));
    }
}